An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{A}$ as shown in the diagram above. The total work done in the process is __________ J.
Explanation:
Work done is given by area enclosed in the P vs V cyclic graph or V vs P cyclic graph.
Sign of work is positive for clockwise cyclic process for V vs P graph.
$\begin{aligned} & W=\frac{1}{2} \times(30-10) \times(30-10)=200 \mathrm{~kPa}-\mathrm{dm}^3 \\ & =200 \times 1000 \mathrm{~Pa}-\mathrm{L}=2 \mathrm{~L}-\mathrm{bar}=200 \mathrm{~J} \end{aligned}$
Standard enthalpy of vapourisation for $\mathrm{CCl}_4$ is $30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Heat required for vapourisation of $284 \mathrm{~g}$ of $\mathrm{CCl}_4$ at constant temperature is ________ $\mathrm{kJ}$.
(Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} ; \mathrm{C}=12, \mathrm{Cl}=35.5$)
Explanation:
$\begin{aligned} & \Delta \mathrm{H}_{\text {vap }}^0 \mathrm{CCl}_4=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \text { Mass of } \mathrm{CCl}_4=284 \mathrm{~gm} \\ & \text { Molar mass of } \mathrm{CCl}_4=154 \mathrm{~g} / \mathrm{mol} \\ & \text { Moles of } \mathrm{CCl}_4=\frac{284}{154}=1.844 \mathrm{~mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1 \mathrm{~mole}=30.5 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta \mathrm{H}_{\text {vap }}{ }^{\circ} \text { for } 1.844 \mathrm{~mol}=30.5 \times 1.844 \\ & \quad=56.242 \mathrm{~kJ} \end{aligned}$
For a certain thermochemical reaction $\mathrm{M} \rightarrow \mathrm{N}$ at $\mathrm{T}=400 \mathrm{~K}, \Delta \mathrm{H}^{\ominus}=77.2 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}=122 \mathrm{~JK}^{-1}, \log$ equilibrium constant $(\log K)$ is __________ $\times 10^{-1}$.
Explanation:
$\begin{aligned} & \Delta \mathrm{G}^{\circ}=\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} \\ & =77.2 \times 10^3-400 \times 122=28400 \mathrm{~J} \\ & \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K} \\ & \Rightarrow 28400=-2.303 \times 8.314 \times 400 \log \mathrm{K} \\ & \Rightarrow \log \mathrm{K}=-3.708=-37.08 \times 10^{-1} \end{aligned}$
If three moles of an ideal gas at $300 \mathrm{~K}$ expand isothermally from $30 \mathrm{~dm}^3$ to $45 \mathrm{~dm}^3$ against a constant opposing pressure of $80 \mathrm{~kPa}$, then the amount of heat transferred is _______ J.
Explanation:
The process involves an ideal gas expanding isothermally, meaning temperature ($T$) remains constant. In an isothermal process for an ideal gas, the change in internal energy ($\Delta U$) is zero:
$\Delta U = 0$ because temperature is constant.
According to the first law of thermodynamics, $\Delta U = Q + W$, where $Q$ is the heat transferred to the system and $W$ is the work done by the system. Since $\Delta U = 0$ in an isothermal process, $Q = -W$.
The work done by the gas during the isothermal expansion against a constant external pressure ($P_{ext}$) is given by:
$W = P_{ext} \times \Delta V$
where:
- $P_{ext}$ = constant opposing pressure = $80$ kPa = $80 \times 10^3$ Pa (since $1$ kPa = $10^3$ Pa)
- $\Delta V$ = change in volume = $45$ dm³ - $30$ dm³ = $15$ dm³ = $15 \times 10^{-3}$ m³ (since $1$ dm³ = $10^{-3}$ m³)
$W = -80 \times 10^3 \times 15 \times 10^{-3} = -1200$ J
Therefore, the amount of heat transferred ($Q$) is $1200$ J, taking into consideration the sign convention that work done by the system is negative.
Explanation:
The entropy change at the melting point is given by $\Delta S = \frac{\Delta H_\mathrm{fus}}{T_\mathrm{m}}$, where $T_\mathrm{m}$ is the melting point in Kelvin. Substituting the given values, we get:
$28.4\ \mathrm{J/K/mol} = \frac{30.4\ \mathrm{kJ/mol}}{T_\mathrm{m}}$
Solving for $T_\mathrm{m}$, we get:
$T_\mathrm{m} = \frac{30.4\ \mathrm{kJ/mol}}{28.4\ \mathrm{J/K/mol}} = 1070.4\ \mathrm{K}$
Rounding off to the nearest integer, the melting point of sodium chloride is $\boxed{1070\ \mathrm{K}}$.
$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB} . \Delta H_{f}^{0}=-200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{AB}, \mathrm{A}_{2}$ and $\mathrm{B}_{2}$ are diatomic molecules. If the bond enthalpies of $\mathrm{A}_{2}, \mathrm{~B}_{2}$ and $\mathrm{AB}$ are in the ratio $1: 0.5: 1$, then the bond enthalpy of $\mathrm{A}_{2}$ is ____________ $\mathrm{kJ} ~\mathrm{mol}^{-1}$ (Nearest integer)
Explanation:
$\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}$
The enthalpy change for the reaction, $\Delta H_{f}^{0}$, is given as:
$\Delta H_{f}^{0} = -200 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ are in the ratio of $1: 0.5: 1$.
Let's denote the bond enthalpies of $\mathrm{A}_{2}$, $\mathrm{B}_{2}$, and $\mathrm{AB}$ as $x$, $0.5x$, and $x$, respectively.
The enthalpy change for the reaction can be calculated using the bond enthalpies:
$\Delta H_{f}^{0} = \text{(Sum of bond enthalpies of reactants)} - \text{(Sum of bond enthalpies of products)}$
For the given reaction:
$-200 = (x + 0.5x) - 2x$
Now we can solve for $x$, which represents the bond enthalpy of $\mathrm{A}_{2}$:
$-200 = 1.5x - 2x$
$-200 = -0.5x$
$x = \frac{-200}{-0.5}$
$x = 400$
The bond enthalpy of $\mathrm{A}_{2}$ is $400 ~\mathrm{kJ} ~\mathrm{mol}^{-1}$.
One mole of an ideal gas at $350 \mathrm{~K}$ is in a $2.0 \mathrm{~L}$ vessel of thermally conducting walls, which are in contact with the surroundings. It undergoes isothermal reversible expansion from 2.0 L to $3.0 \mathrm{~L}$ against a constant pressure of $4 \mathrm{~atm}$. The change in entropy of the surroundings ( $\Delta \mathrm{S})$ is ___________ $\mathrm{J} \mathrm{K}^{-1}$ (Nearest integer)
Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
Explanation:
For an isothermal, reversible process, the change in entropy (ΔS) of the system can be calculated using the formula:
$ \Delta S = nR \ln\left(\frac{V_2}{V_1}\right) $
where (n) is the number of moles, (R) is the gas constant, and (V_2) and (V_1) are the final and initial volumes, respectively.
Substituting the given values:
$ \Delta S_{\text{system}} = 1 \times 8.314 \, \ln\left(\frac{3}{2}\right) = 3.37 \, \text{J K}^{-1} $
Since the process is reversible and the total entropy change in the universe should be zero for reversible processes, the change in entropy of the surroundings is equal to the negative change of the system's entropy:
$ \Delta S_{\text{surroundings}} = - \Delta S_{\text{system}} = -3.37 \, \text{J K}^{-1} $
But considering the heat transfer from the system to the surroundings, the sign should be positive, which means the entropy of the surroundings also increases:
$ \Delta S_{\text{surroundings}} = 3.37 \, \text{J K}^{-1} $
So, rounded to the nearest integer,
$ \Delta S_{\text{surroundings}} = 3 \, \text{J K}^{-1} $
The total number of intensive properties from the following is __________
Volume, Molar heat capacity, Molarity, $\mathrm{E}^{\theta}$ cell, Gibbs free energy change, Molar mass, Mole
Explanation:
Intensive properties :
1. Molarity : Molarity is a measure of concentration, defined as the number of moles of solute per liter of solution. It doesn't depend on the total volume of the solution but only on the ratio of the amount of solute to the amount of solution.
2. $\mathrm{E}^{\theta}$ cell (standard cell potential) : This is a measure of the potential difference between the anode and cathode in an electrochemical cell under standard conditions. It's independent of the amount of material in the cell.
3. Molar heat capacity : This is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or Kelvin). It's an intrinsic property because it's defined per mole of substance.
4. Molar mass : Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's independent of the quantity of the substance because it's defined per mole.
On the other hand, extensive properties depend on the amount of the substance. They change when the size or mass of a sample changes. For instance, mass and volume are extensive properties.
Extensive properties :
1. Volume: The volume of a substance depends on the amount of that substance. For instance, one liter of water has half the volume of two liters of water.
2. Gibbs free energy change (∆G) : This is the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It depends on the number of moles of reactants and products, so it's an extensive property.
3. Mole : The mole is a unit of measurement for amount of substance in the International System of Units (SI). It's a count of a very large number of particles, typically atoms or molecules. Like mass or volume, it's an extensive property because it depends on the quantity of the substance.
Solid fuel used in rocket is a mixture of $\mathrm{Fe}_{2} \mathrm{O}_{3}$ and $\mathrm{Al}$ (in ratio 1 : 2). The heat evolved $(\mathrm{kJ})$ per gram of the mixture is ____________. (Nearest integer)
Given: $\Delta \mathrm{H}_{\mathrm{f}}^{\theta}\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)=-1700 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta \mathrm{H}_{\mathrm{f}}^{\theta}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)=-840 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Molar mass of Fe, Al and O are 56, 27 and 16 g mol$^{-1}$ respectively.
Explanation:
First, let's consider the reaction :
$2\mathrm{Al}(s) + \mathrm{Fe}_2\mathrm{O}_3(s) \rightarrow \mathrm{Al}_2\mathrm{O}_3(s) + 2\mathrm{Fe}(s)$
The heat change for this reaction $\Delta H^0$ can be calculated from the heats of formation of the reactants and the products :
$\Delta H^0 = [\Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) + 2\Delta H_f^0(\mathrm{Fe})] - [2\Delta H_f^0(\mathrm{Al}) + \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)]$
Assuming the elements in their standard states have zero enthalpy of formation, i.e.,
$\Delta H_f^0(\mathrm{Al}) = \Delta H_f^0(\mathrm{Fe}) = 0$, we can simplify this to :
$\Delta H^0 = \Delta H_f^0(\mathrm{Al}_2\mathrm{O}_3) - \Delta H_f^0(\mathrm{Fe}_2\mathrm{O}_3)$
Substitute the given heats of formation into the equation :
$\Delta H^0 = (-1700\ \mathrm{kJ/mol}) - (-840\ \mathrm{kJ/mol}) = -860\ \mathrm{kJ/mol}$
This is the heat of reaction for the above reaction. However, we are asked to find the heat evolved per gram of the mixture.
To find this, we need to determine the molar mass of the reactants in the reaction. The molar mass of $\mathrm{Fe}_2\mathrm{O}_3$ is $2 \times 56 + 3 \times 16 = 160\ \mathrm{g/mol}$ and the molar mass of 2 moles of $\mathrm{Al}$ is $2 \times 27 = 54\ \mathrm{g/mol}$. The total molar mass of the mixture is $160 + 54 = 214\ \mathrm{g/mol}$.
So, the heat evolved per gram of the mixture is $\frac{-860\ \mathrm{kJ/mol}}{214\ \mathrm{g/mol}} = -4.0187\ \mathrm{kJ/g}$
Rounded to the nearest integer, this value is approximately -4 kJ/g. The negative sign indicates that the heat is evolved (exothermic reaction).
The number of endothermic process/es from the following is ______________.
A. $\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{I}(\mathrm{g})$
B. $\mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})$
C. $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
D. $\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$
E. Dissolution of ammonium chloride in water
Explanation:
An endothermic process is one that absorbs heat from the surroundings. Here are the types of the given reactions:
A. $\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{I}(\mathrm{g})$
This process involves the dissociation of iodine molecules into individual iodine atoms. This requires energy to break the bonds between the iodine atoms, so it is an endothermic process.
B. $\mathrm{HCl}(\mathrm{g}) \rightarrow \mathrm{H}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})$
This is the dissociation of hydrogen chloride into hydrogen and chlorine. Similar to the first reaction, it requires energy to break the bond between the hydrogen and chlorine atoms. Therefore, this is also an endothermic process.
C. $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$
This is the process of evaporation or vaporization, where liquid water is converted to water vapor. Evaporation is an endothermic process because it requires heat to break the hydrogen bonds between water molecules and convert it into a gas.
D. $\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})$
This is a combustion reaction. Combustion reactions are generally exothermic because they release energy in the form of heat and light. So, this is not an endothermic process.
E. Dissolution of ammonium chloride in water
The dissolution of ammonium chloride ($\mathrm{NH}_4\mathrm{Cl}$) in water is known to be endothermic. The ionic bonds in the ammonium chloride and the hydrogen bonds in the water must be broken for the salt to dissolve, which requires heat.
So, in summary, four of these processes (A, B, C, and E) are endothermic.
For complete combustion of ethene.
$\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
the amount of heat produced as measured in bomb calorimeter is $1406 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $300 \mathrm{~K}$. The minimum value of $\mathrm{T} \Delta \mathrm{S}$ needed to reach equilibrium is ($-$) _________ $\mathrm{kJ}$. (Nearest integer)
Given : $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
Explanation:
In the given combustion reaction, the number of moles of gaseous products ($\Delta n_g$) is -2 (since we have 2 moles of COâ‚‚ gas on the product side and 4 moles of gaseous reactants).
The change in internal energy ($\Delta U$) is given as -1406 kJ/mol.
We can calculate the change in enthalpy ($\Delta H$) using the equation $\Delta H = \Delta U + \Delta n_gRT$, where $R$ is the ideal gas constant and $T$ is the temperature.
Substituting the given values:
$ \begin{aligned} \Delta H &= \Delta U + \Delta n_gRT \\\\ &= -1406 \mathrm{kJ/mol} + (-2) \times 8.3 \times 10^{-3} \mathrm{kJ/K/mol} \times 300 \mathrm{K} \\\\ &= -1406 \mathrm{kJ/mol} - 4.98 \mathrm{kJ/mol} \\\\ &\approx -1411 \mathrm{kJ/mol} \end{aligned} $
At equilibrium, $\Delta G = 0$, so $\Delta H = T \Delta S$, which gives $T \Delta S = \Delta H$.
Therefore, the minimum value of $T \Delta S$ needed to reach equilibrium is -1411 kJ
When a $60 \mathrm{~W}$ electric heater is immersed in a gas for 100 s in a constant volume container with adiabatic walls, the temperature of the gas rises by $5^{\circ} \mathrm{C}$. The heat capacity of the given gas is ___________ $\mathrm{J} \mathrm{K}^{-1}$ (Nearest integer)
Explanation:
The heat provided by the heater is given by the equation:
$Q = \text{Power} \times \text{Time}$
Substituting the given values:
$Q = 60 \, \text{W} \times 100 \, \text{s} = 6000 \, \text{J}$
The heat capacity (C) is defined as the amount of heat required to raise the temperature of a substance by one degree. It is given by the equation:
$C = Q/\Delta T$
Substituting the given values:
$C = 6000 \, \text{J} / 5 \, \text{°C} = 1200 \, \text{J/K}$
So, the heat capacity of the given gas is approximately 1200 J/K.
Consider the following data
Heat of combustion of $\mathrm{H}_{2}(\mathrm{g})\quad\quad=-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Heat of combustion of $\mathrm{C}(\mathrm{s})\quad\quad=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Heat of combustion of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})\quad=-1234.7 \mathrm{~kJ}~{\mathrm{mol}}^{-1}$
The heat of formation of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ is $(-)$ ___________ $\mathrm{kJ} ~\mathrm{mol}^{-1}$ (Nearest integer).
Explanation:
$2\mathrm{C}(\mathrm{s}) + 6\mathrm{H}_{2}(\mathrm{g}) + \frac{1}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$
The heat of combustion for $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ is:
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$
By adding the formation reaction with the combustion reaction times -1, we get:
$2\mathrm{C}(\mathrm{s}) + 3\mathrm{H}_{2}(\mathrm{g}) + \frac{3}{2}\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2\mathrm{CO}_{2}(\mathrm{g}) + 3\mathrm{H}_{2}\mathrm{O}(\mathrm{l})$
This reaction has a heat equal to the heat of formation of ethanol times -1.
Therefore, the heat of formation of ethanol is equal to:
$-1234.7 \mathrm{~kJ/mol} + 2*(-393.5 \mathrm{~kJ/mol}) + 3*(-241.8 \mathrm{~kJ/mol})$
Calculating this gives:
$\Delta H_f(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})) = -277.7 \approx -278 \mathrm{~kJ/mol}$ (rounded to the nearest integer)
Consider the graph of Gibbs free energy G vs Extent of reaction. The number of statement/s from the following which are true with respect to points (a), (b) and (c) is _________
A. Reaction is spontaneous at (a) and (b)
B. Reaction is at equilibrium at point (b) and non-spontaneous at point (c)
C. Reaction is spontaneous at (a) and non-spontaneous at (c)
D. Reaction is non-spontaneous at (a) and (b)
Explanation:
For, Equilibrium $\mathrm{dG}=0$
For, Nonspontaneous process $\mathrm{dG}>0$
$\therefore $ A Wrong
B Correct
C Correct
D Wrong
The value of $\log \mathrm{K}$ for the reaction $\mathrm{A} \rightleftharpoons \mathrm{B}$ at $298 \mathrm{~K}$ is ___________. (Nearest integer)
Given: $\Delta \mathrm{H}^{\circ}=-54.07 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\Delta \mathrm{S}^{\circ}=10 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$
(Take $2.303 \times 8.314 \times 298=5705$ )
Explanation:
Given:
$
\begin{align}
\Delta H^0 & = -54.07 \, \text{kJ/mol} = -54070 \, \text{J/mol} \\\\
\Delta S^0 & = 10 \, \text{J/K}\cdot \text{mol} \\\\
T & = 298 \, \text{K}
\end{align}
$
We find the change in Gibbs free energy $\Delta G^0$:
$
\begin{align}
\Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\
& = -54070 - (10 \times 298) \\\\
& = -54070 - 2980 \\\\
& = -57050 \, \text{J/mol}
\end{align}
$
Now, we'll use the given expression with the correct constant for $ R $ as 8.314 J/(mol·K):
$
\begin{align}
\Delta G^0 & = -2.303 \times 8.314 \times 298 \times \log K \\\\
\log K & = \frac{-57050}{2.303 \times 8.314 \times 298} \\\\
& \approx 10
\end{align}
$
So, the answer is $\log K = 10$.
$0.3 \mathrm{~g}$ of ethane undergoes combustion at $27^{\circ} \mathrm{C}$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $0.5^{\circ} \mathrm{C}$. The heat evolved during combustion of ethane at constant pressure is ____________ $\mathrm{kJ} ~\mathrm{mol}{ }^{-1}$. (Nearest integer)
[Given : The heat capacity of the calorimeter system is $20 \mathrm{~kJ} \mathrm{~K}^{-1}, \mathrm{R}=8.3 ~\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
Assume ideal gas behaviour.
Atomic mass of $\mathrm{C}$ and $\mathrm{H}$ are 12 and $1 \mathrm{~g} \mathrm{~mol}^{-1}$ respectively]
Explanation:
$\begin{aligned} & \text { No. of moles of ethane }=\frac{0.3}{30}=0.01 \\\\ & \text { Heat evolved in Bomb calorimeter }=20 \times 0.5 \\\\ & =10 \mathrm{~kJ} \\\\ & \Delta \mathrm{U}=-\frac{10}{0.01}=-1000 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ & =-1000+(-2.5) \times \frac{8.3 \times 300}{1000} \\\\ & =-1000-6.225 \\\\ & =-1006.225 \\\\ & |\Delta \mathrm{H}| \simeq 1006 \mathrm{~kJ} \mathrm{~mol}^{-1} \\\\ & \end{aligned}$
At $25^{\circ} \mathrm{C}$, the enthalpy of the following processes are given :
| $\mathrm{H_2(g)+O_2(g)}$ | $\to$ | $2\mathrm{OH(g)}$ | $\mathrm{\Delta H^\circ=78~kJ~mol^{-1}}$ |
|---|---|---|---|
| $\mathrm{H_2(g)+\frac{1}{2}O_2(g)}$ | $\to$ | $\mathrm{H_2O(g)}$ | $\mathrm{\Delta H^\circ=-242~kJ~mol^{-1}}$ |
| $\mathrm{H_2(g)}$ | $\to$ | $\mathrm{2H(g)}$ | $\mathrm{\Delta H^\circ=436~kJ~mol^{-1}}$ |
| $\frac{1}{2}\mathrm{O_2(g)}$ | $\to$ | $\mathrm{O(g)}$ | $\mathrm{\Delta H^\circ=249~kJ~mol^{-1}}$ |
What would be the value of X for the following reaction ? _____________ (Nearest integer)
$\mathrm{H_2O(g)\to H(g)+OH(g)~\Delta H^\circ=X~kJ~mol^{-1}}$
Explanation:
$ \begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\frac{436+78}{2}-(-242) \\\\ & =\frac{436+78}{2}+242=499 \end{aligned} $
$\mathrm{CCl}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})$
Explanation:
Enthalpy of above reaction
$ \begin{aligned} & =\Delta \mathrm{H}_f\left(\mathrm{CO}_2(\mathrm{~g})\right)+4 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl}(\mathrm{g}))-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CCl}_4\right)-2 \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right) \\\\ & =-394-4 \times 92+105+2 \times 242 \\\\ & =-394-368+105+484 \\\\ & =-173 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
Hence the magnitude of this will be $173 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
The enthalpy change for the conversion of $\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})$ to $\mathrm{Cl}^{-}$(aq) is ($-$) ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer)
Given : $\Delta_{\mathrm{dis}} \mathrm{H}_{\mathrm{Cl}_{2(\mathrm{~g})}^{\theta}}^{\ominus}=240 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta_{\mathrm{eg}} \mathrm{H}_{\mathrm{Cl_{(g)}}}^{\ominus}=-350 \mathrm{~kJ} \mathrm{~mol}^{-1}$,
${\mathrm{\Delta _{hyd}}H_{Cl_{(g)}^ - }^\Theta = - 380}$ $\mathrm{kJ~mol^{-1}}$
Explanation:
Given $\mathrm{C_V}=20 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Explanation:
$\mathrm{T_1=300~K}$
$\mathrm{w = 3}$ kJ/mole
$\mathrm{w=nC_v\Delta T}$
$3000=1\times20\times(300-\mathrm{T_2})$
$300-\mathrm{T_2}=150$
$\mathrm{T_2=150~K}$
When 2 litre of ideal gas expands isothermally into vacuum to a total volume of 6 litre, the change in internal energy is ____________ J. (Nearest integer)
Explanation:
For isothermal process of an ideal gas; $\mathrm{\Delta E=0}$
28.0 L of CO$_2$ is produced on complete combustion of 16.8 L gaseous mixture of ethene and methane at 25$^\circ$C and 1 atm. Heat evolved during the combustion process is ___________ kJ.
Given : $\mathrm{\Delta H_c~(CH_4)=-900~kJ~mol^{-1}}$
$\mathrm{\Delta H_c~(C_2H_4)=-1400~kJ~mol^{-1}}$
Explanation:
Let, Volume of C2H4 is x litre
$ \begin{aligned} &\Rightarrow 28=16.8+\mathrm{x} \\\\ &\mathrm{x}=11.2 \mathrm{~L} \\\\ &\mathrm{n}_{\mathrm{CH}_4}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1 \times 5.6}{0.082 \times 298}=0.229 \text { mole } \\\\ &\mathrm{n}_{\mathrm{C}_2 \mathrm{H}_2}=\frac{11.2}{0.082 \times 298}=0.458 \text { mole } \\\\ &\therefore \text { Heat evolved }=0.229 \times 900+0.458 \times 1400 \\\\ &=206.1+641.2 \\\\ &=847.3 \mathrm{~kJ} \end{aligned} $
An athlete is given 100 g of glucose (C$_6$H$_{12}$O$_6$) for energy. This is equivalent to 1800kJ of energy. The 50% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ____________ g (Nearest integer)
Assume that there is no other way of consuming stored energy.
Given : The enthalpy of evaporation of water is 45 kJ mol$^{-1}$
Molar mass of C, H & O are 12, 1 and 16 g mol$^{-1}$.
Explanation:
Extra energy used to convert $\mathrm{H}_2 \mathrm{O(l)}$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ into $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
$ \begin{aligned} & =\frac{1800}{2}=900 \mathrm{~kJ} \\\\ & \Rightarrow 900=\mathrm{n}_{\mathrm{H}_2 \mathrm{O}} \times 45 \\\\ & \mathrm{n}_{\mathrm{H}_2 \mathrm{O}}=\frac{900}{45}=20 \text { mole } \\\\ & \mathrm{W}_{\mathrm{H}_2 \mathrm{O}}=20 \times 18=360 \mathrm{~g} \end{aligned} $
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is __________ J (nearest integer)
Given ; $\log2=0.3$
$\ln10=2.3$
Explanation:
$2 \rightarrow 3 \Rightarrow$ Isochoric process
$3 \rightarrow 1 \Rightarrow$ Isothermal process
$\mathrm{W}=\mathrm{W}_{1 \rightarrow 2}+\mathrm{W}_{2 \rightarrow 3}+\mathrm{W}_{3 \rightarrow 1}$
$=\left(-\mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right)+0\left[-\mathrm{P}_1 \mathrm{~V}_1 \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)\right]\right)$
$=\left[-1 \times(40-20)+0+\left[-1 \times 20 \ln \left(\frac{20}{40}\right)\right]\right]$
$=-20+20 \ln 2$
$=-20+20 \times 2.3 \times 0.3$
$=-6.2$ bar $L$
$|\mathrm{W}|=6.2$ bar $1=620 \mathrm{~J}$
Following figure shows spectrum of an ideal black body at four different temperatures. The number of correct statement/s from the following is ____________.
A. $\mathrm{T_4 > T_3 > T_2 > T_1}$
B. The black body consists of particles performing simple harmonic motion.
C. The peak of the spectrum shifts to shorter wavelength as temperature increases.
D. ${{{T_1}} \over {{v_1}}} = {{{T_2}} \over {{v_2}}} = {{{T_3}} \over {{v_3}}} \ne $ constant
E. The given spectrum could be explained using quantisation of energy.
Explanation:
B. It is incorrect as particles do not undergo simple harmonic motion.
C. It is correct
D. It is incorrect
E. It is correct
For independent process at 300 K
| Process | $\mathrm{\Delta H/kJ~mol^{-1}}$ | $\mathrm{\Delta S/J~K^{-1}}$ |
|---|---|---|
| A | $-25$ | $-80$ |
| B | $-22$ | 40 |
| C | 25 | $-50$ |
| D | 22 | 20 |
The number of non-spontaneous process from the following is __________
Explanation:
Processes $\mathrm{C}$ and $\mathrm{D}$ are non-spontaneous.
When 600 mL of 0.2 M HNO3 is mixed with $400 \mathrm{~mL}$ of 0.1 M NaOH solution in a flask, the rise in temperature of the flask is ___________ $\times 10^{-2}{ }\,^{\circ} \mathrm{C}$.
(Enthalpy of neutralisation $=57 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and Specific heat of water $=4.2 \,\mathrm{JK}^{-1} \mathrm{~g}^{-1}$) (Neglect heat capacity of flask)
Explanation:
600 mL × 0.2 M = 120 m mol
NaOH
400 mL × 0.1 M = 40 m mol
Heat liberated from reaction
$ =40 \times 10^{-3} \times 57 \times 10^{3} \mathrm{~J} $
Heat gained by solution $=m C \Delta T$
$ \begin{aligned} \mathrm{m}=\text { mass of solution }=\mathrm{V} \times \mathrm{d} &=1000 \times 1 \\\\ &=1000 \mathrm{~g} \end{aligned} $
Heat gained by solution $=1000 \times 4.2 \times \Delta \mathrm{T} \ldots(2)$
From (1) and (2)
Heat liberated $=$ Heat gained
$40 \times 10^{-3} \times 57 \times 10^{3}=1000 \times 4.2 \times \Delta T$
$\Delta T=54 \times 10^{-2}{ }^{\circ} \mathrm{C}$
(Rounded off to the nearest integer)
Among the following the number of state variables is ______________.
Internal energy (U)
Volume (V)
Heat (q)
Enthalpy (H)
Explanation:
A gas (Molar mass = 280 $\mathrm{~g} \mathrm{~mol}^{-1}$) was burnt in excess $\mathrm{O}_{2}$ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from $298.0 \mathrm{~K}$ to $298.45$ $\mathrm{K}$. If the heat capacity of calorimeter is $2.5 \mathrm{~kJ} \mathrm{~K}^{-1}$ and enthalpy of combustion of gas is $9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ then amount of gas burnt is _____________ g. (Nearest Integer)
Explanation:
$ \begin{aligned} &=2.5 \times 10^{3} \times 0.45 \\\\ &=1.125 \mathrm{~kJ} \end{aligned} $
Considering $\Delta \mathrm{H} \simeq \Delta \mathrm{U}$
$ \Delta \mathrm{H}=9 \mathrm{~kJ} / \mathrm{mol} \simeq \Delta \mathrm{U} $
$\therefore$ Mass of gas burnt $=\frac{1.125}{9} \times 280=35 \mathrm{~g}$
The molar heat capacity for an ideal gas at constant pressure is $20.785 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$. The change in internal energy is $5000 \mathrm{~J}$ upon heating it from $300 \mathrm{~K}$ to $500 \mathrm{~K}$. The number of moles of the gas at constant volume is ____________. [Nearest integer] (Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$ \begin{aligned} &\text { and } \Delta \mathrm{U}=\mathrm{nC} \mathrm{v} \Delta \mathrm{T} \\\\ &\therefore \quad \mathrm{nC}_{\mathrm{v}}=\frac{5000}{200}=25 \end{aligned} $
and we know that
$ \begin{aligned} &C_{p}-C_{v}=R \\\\ &20.785-\frac{25}{n}=8.314 \\\\ &n=\frac{25}{(20.785-8.314)}=2 \end{aligned} $
For the reaction
$\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$
$\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$.
The enthalpy change for the above reaction is ($-$) __________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$ [nearest integer]
Given: $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
Explanation:
$\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$
$ \begin{aligned} \Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\ &=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\ &=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$2.4 \mathrm{~g}$ coal is burnt in a bomb calorimeter in excess of oxygen at $298 \mathrm{~K}$ and $1 \mathrm{~atm}$ pressure. The temperature of the calorimeter rises from $298 \mathrm{~K}$ to $300 \mathrm{~K}$. The enthalpy change during the combustion of coal is $-x \mathrm{~kJ} \mathrm{~mol}^{-1}$. The value of $x$ is ___________. (Nearest Integer)
(Given : Heat capacity of bomb calorimeter $20.0 \mathrm{~kJ} \mathrm{~K}^{-1}$. Assume coal to be pure carbon)
Explanation:
$\mathrm{n}_{\text {coal }}=\frac{2.4}{12}$
$\mathrm{Q}=\frac{-20(300-298)}{0.2}$
$Q=-200 \mathrm{~kJ} / \mathrm{mol}$
$x=200$
While performing a thermodynamics experiment, a student made the following observations.
HCl + NaOH $\to$ NaCl + H2O $\Delta$H = $-$57.3 kJ mol$-$1
CH3COOH + NaOH $\to$ CH3COONa + H2O $\Delta$H = $-$55.3 kJ mol$-$1
The enthalpy of ionization of CH3COOH as calculated by the student is _____________ kJ mol$-$1. (nearest integer)
Explanation:
$ \Delta \mathrm{H}_{1}=-57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
(II) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{H}_{2} \mathrm{O}$
$ \Delta \mathrm{H}_{2}=-55.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
Reaction (I) can be written as
$ \text { (III) } \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HCl}+\mathrm{NaOH} $
$ \Delta \mathrm{H}_{3}=57.3 \,\mathrm{KJ} \mathrm{mol}^{-1} $
By adding (II) and (III)
$ \begin{aligned} &\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{NaCl} \rightarrow \mathrm{CH}_{3} \mathrm{COONa}+\mathrm{HCl} \quad \Delta \mathrm{H}_{\mathrm{r}} \\ &\begin{aligned} \Delta \mathrm{H}_{\mathrm{r}}=\Delta \mathrm{H}_{3}+\Delta \mathrm{H}_{2} &=57.3-55.3 \\ &=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} \end{aligned} $
The enthalpy of combustion of propane, graphite and dihydrogen at $298 \mathrm{~K}$ are $-2220.0 \mathrm{~kJ} \mathrm{~mol}^{-1},-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The magnitude of enthalpy of formation of propane $\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)$ is _______________ $\mathrm{kJ} \,\mathrm{mol}^{-1}$. (Nearest integer)
Explanation:
$\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \Delta \mathrm{H}_{1}=-2220 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{C}($ graphite $)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g}), \quad \Delta \mathrm{H}_{2}=-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{I}), \quad \Delta \mathrm{H}_{3}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The desired reaction is
$3 \mathrm{C}$ (graphite) + $4 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$
$\Delta \mathrm{H}_{\mathrm{f}}=3 \Delta \mathrm{H}_{2}+4 \Delta \mathrm{H}_{3}-\Delta \mathrm{H}_{1}$
$ \begin{aligned} &=3(-393.5)+4(-285.8)-(-2220) \\ &=-103.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
$\left|\Delta \mathrm{H}_{\mathrm{f}}\right| \simeq 104 \mathrm{~kJ} \mathrm{~mol}^{-1}$
1.0 mol of monoatomic ideal gas is expanded from state 1 to state 2 as shown in the figure. The magnitude of the work done for the expansion of gas from state 1 to state 2 at 300 K is ____________ J. (Nearest integer)
(Given : R = 8.3 J K$-$1 mol$-$1, ln10 = 2.3, log2 = 0.30)
Explanation:
2.2 g of nitrous oxide (N2O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, $\Delta$U is '$-$x' J. The value of 'x' is ________. [nearest integer]
(Given : atomic mass of N = 14 g mol$-$1 and of O = 16 g mol$-$1. Molar heat capacity of N2O is 100 J K$-$1 mol$-$1)
Explanation:
$ \begin{aligned} \Delta U &=q+w \\\\ &=\frac{100 \times 2.2}{44}(-40)-(-49.39) \times 10^{-3} \times 101.325 \end{aligned} $
$ \begin{aligned} &=-200+5 \\\\ &=-195 \mathrm{~J} \\\\ \mathrm{x}=& 195 \end{aligned} $
17.0 g of NH3 completely vapourises at $-$33.42$^\circ$C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol$-$1. The enthalpy change for the vapourisation of 85 g of NH3 under the same conditions is _________ kJ.
Explanation:
So, required $\Delta \mathrm{H}=5 \times 23.4$
$ =117 \mathrm{~kJ} $
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\Delta$CH$\Theta $ = $-$601.70 kJ mol$-$1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K$-$1 mol$-$1)
Explanation:
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\ &\Delta \mathrm{ng}=-\frac{1}{2} \\\\ &-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\ &\Delta \mathrm{U}=-601.70+1.245 \\\\ &\Delta \mathrm{U} \simeq-600 \mathrm{~kJ} \end{aligned} $
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 2.0 L. The amount of heat absorbed in this expansion is ____________ L atm.
Explanation:
$ \because P_{\text {ext }}=0 \quad \text { (vacuum) } $
$\therefore w=0, \quad\Delta U=0$ (as the process is isothermal)
So, $q=0$
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is __________ J. [nearest integer] (Given : R = 8.3 J K$-$1 mol$-$1 and log 2 = 0.3010)
Explanation:
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100$^\circ$C, then the internal energy for vaporization in kJ mol$-$1 is ___________. [nearest integer]
[Assume steam to be an ideal gas. Given $\Delta$vapH$^\Theta $ for water at 373 K and 1 bar is 41.1 kJ mol$-$1 ; R = 8.31 J K$-$1 mol$-$1]
Explanation:
$ \begin{aligned} \mathrm{n}_{\mathrm{H}_{2} \mathrm{O}} &=\frac{36}{18}=2 \quad \Delta \mathrm{n}_{\mathrm{g}}=1-0=1 \\\\ \Delta \mathrm{U}_{\text {vap }} &=\Delta \mathrm{H}_{\text {vap }}-\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=41.1-(1) \times 8.31 \times 10^{-3} \times 373 \\\\ &=41.1-3.099 \\\\ &=38 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $
For complete combustion of methanol
CH3OH(I) + ${3 \over 2}$O2(g) $\to$ CO2(g) + 2H2O(I)
the amount of heat produced as measured by bomb calorimeter is 726 kJ mol$-$1 at 27$^\circ$C. The enthalpy of combustion for the reaction is $-$x kJ mol$-$1, where x is ___________. (Nearest integer)
(Given : R = 8.3 JK$-$1 mol$-$1)
Explanation:
$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_{\mathrm{g}} \mathrm{RT} \\\\ &=-726 \mathrm{~kJ}+\left(\frac{-1}{2}\right) \times 8.3 \times 300 \\\\ &\simeq-727 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
The standard entropy change for the reaction
4Fe(s) + 3O2(g) $\to$ 2Fe2O3(s) is $-$550 J K$-$1 at 298 K.
[Given : The standard enthalpy change for the reaction is $-$165 kJ mol$-$1]. The temperature in K at which the reaction attains equilibrium is _____________. (Nearest Integer)
Explanation:
$ \begin{aligned} &\Rightarrow-165 \times 10^3-\mathrm{T} \times(-505)=0 \\\\ &\Rightarrow \mathrm{T}=300 \mathrm{~K} \end{aligned} $
Explanation:
$\Delta$S = $-$ 176 JK$-$1 mol$-$1
T = 298 K
Using Gibb's free energy relation
$\Delta$G = $\Delta$H $-$ T$\Delta$S
where, $\Delta$G = change in Gibb's free energy
$\Delta$H = change in enthalpy
T = temperature
$\Delta$S = change in entropy
$\Delta$G = 57.8 kJ/mol $-$ [298 K $\times$ ($-$ 176 Jk$-$1 mol$-$1)]
= 57.8 kJ/mol $-$ $\left( {298 \times {{ - 176} \over {1000}}kJ} \right)$ [$\therefore$ 1 kJ = 1000 J]
= $-$ 5.352 kJ/mol
| $\Delta$G | = 5.352
Hence, answer is 5.
When the valve is opened, the final pressure of the system in bar is x $\times$ 10$-$2. The value of x is __________. (Integer answer)
[Assume - Ideal gas; 1 bar = 105 Pa; Molar mass of N2 = 28.0 g mol$-$1; R = 8.31 J mol$-$1 K$-$1]
Explanation:
$\Rightarrow$ Assuming the system attains a final temperature of T (such that 300 < T < 60)
$\Rightarrow$ $\left( {\matrix{ {Heat\,lost\,by} \cr {{N_2}\,of\,container} \cr I \cr } } \right) = \left( {\matrix{ {Heat\,gained\,by} \cr {{N_2}\,of\,container} \cr {II} \cr } } \right)$
$\Rightarrow$ n1Cm(300 $-$ T) = nIICm(T $-$ 60)
$ \Rightarrow \left( {{{2.8} \over {28}}} \right)(300 - T) = {{0.2} \over {28}}(T - 60)$
$\Rightarrow$ 14(300 $-$ T) = T $-$ 60
$\Rightarrow$ ${{(14 \times 300 + 60)} \over {15}} = T$
$\Rightarrow$ T = 284 K (final temperature)
$\Rightarrow$ If the final pressure = P
$\Rightarrow$ (nI + nII)final = $\left( {{{3.0} \over {28}}} \right)$
$\Rightarrow$${P \over {RT}}({V_I} + {V_{II}}) = {{3.0\,gm} \over {28\,gm/mol}}$
$P = \left( {{3 \over {28}}mol} \right) \times 8.31{J \over {mol - K}} \times {{284K} \over {3 \times {{10}^{ - 3}}{m^3}}} \times {10^{ - 5}}{{bar} \over {Pa}}$
$\Rightarrow$ 0.84287 bar
$\Rightarrow$ 84.28 $\times$ 10$-$2 bar
$\Rightarrow$ 84
FeO(s) + C(graphite) $\to$ Fe(s) + CO(g)
| Substance | $\Delta H^\circ $ (kJ mol$^{ - 1}$) |
$\Delta S^\circ $ (J mol$^{ - 1}$ K$^{ - 1}$) |
|---|---|---|
| $Fe{O_{(s)}}$ | $ - 266.3$ | 57.49 |
| ${C_{(graphite)}}$ | 0 | 5.74 |
| $F{e_{(s)}}$ | 0 | 27.28 |
| $C{O_{(g)}}$ | $ - 110.5$ | 197.6 |
The minimum temperature in K at which the reaction becomes spontaneous is ___________. (Integer answer)
Explanation:
${\Delta ^0}{H_{rxn}} = \left[ {\Delta _f^0H(Fe) + \Delta _f^0H(CO)} \right] - $
$ = \left[ {\Delta _f^0H(FeO) + \Delta _f^0H({C_{(graphite)}})} \right]$
$ = [0 - 110.5] - [ - 266.3 + 0]$
= 155.8 kJ/mol
${\Delta ^0}{S_{rxn}} = \left[ {{\Delta ^0}S(Fe) + {\Delta ^0}S(CO)} \right] - $
$\left[ {{\Delta ^0}S(FeO) + {\Delta ^0}S({C_{(graphite)}})} \right]$
$ = [27.28 + 197.6] - [57.49 + 5.74]$
= 161.65 J/mol-K
${T_{\min }} = {{155.8 \times {{10}^3}J/mol} \over {161.65J/mol - K}} = 963.8$ K
$ \simeq 964$ k (nearest integer)
[Given : Specific heat of water = 4.18 J g$-$1 K$-$1, Density of water = 1.00 g cm$-$3]
[Assume no volume change on mixing)
Explanation:
$\Rightarrow$ Millimoles of NaOH = 300 $\times$ 0.1 = 30
$\Rightarrow$ Heat released = $\left( {{{30} \over {1000}} \times 57.1 \times 1000} \right)$ = 1713 J
$\Rightarrow$ Mass of solution = 500 ml $\times$ 1 gm/ml = 500 gm
$\Rightarrow$ $\Delta T = {q \over {m \times c}} = {{1713J} \over {500g \times 4.18{J \over {g - K}}}}$ = 0.8196 K
= 81.96 $\times$ 10$-$2 K
[Use : R = 8.3 J mol$-$1 K$-$1]
Explanation:
$\Rightarrow$ From the relation : $\Delta$H = $\Delta$U + $\Delta$ngRT
$\Rightarrow$ 41${{kJ} \over {mol}}$ = $\Delta$U + (1) $\times$ ${{8.3} \over {1000}}$ $\times$ 373
$\Delta$ DU = 41 $-$ 3.0959 = 38 kJ/mol
${\Delta _f}{H^\Theta }$ for KCl = $-$436.7 kJ mol$-$1 ;
${\Delta _{sub}}{H^\Theta }$ for K = 89.2 kJ mol$-$1 ;
${\Delta _{ionization}}{H^\Theta }$ for K = 419.0 kJ mol$-$1 ;
${\Delta _{electron\,gain}}{H^\Theta }$ for Cl(g) = $-$348.6 kJ mol$-$1 ;
${\Delta _{bond}}{H^\Theta }$ for Cl2 = 243.0 kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is _____________ (Nearest integer)
Explanation:
$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $
$ \Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$ kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is 718 (Nearest integer).
[Use : H+ (aq) + OH$-$ (aq) $\to$ H2O : $\Delta$$\gamma$H = $-$57.1 kJ mol$-$1]
Specific heat of H2O = 4.18 J K$-$1 g$-$1
density of H2O = 1.0 g cm$-$3
Assume no change in volume of solution on mixing.
Explanation:
${n_{O{H^ - }}} = {{600 \times 0.1} \over {1000}} = 0.06$ (L.R.)
Now, heat liberated from reaction = heat gained by solutions
or, 0.06 $\times$ 57.1 $\times$ 103
= (1000 $\times$ 1.0) $\times$ 4.18 $\times$ $\Delta$T
$\therefore$ $\Delta$T = 0.8196K
= 81.96 $\times$ 10$-$2 K $ \approx $ 82 $\times$ 10$-$2 K