Thermodynamics
At $300 \mathrm{~K}, \Delta_r G^{\Theta}$ for the reaction $A_2(g) \rightleftharpoons B_2(g)$ is $-11.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The Equilibrium constant at 300 K is approximately ( $R=8314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ )
Two statements are given below.
Statement I : The reaction $\mathrm{Cr}_2 \mathrm{O}_3+2 \mathrm{Al} \longrightarrow \mathrm{Al}_2 \mathrm{O}_3+2 \mathrm{Cr}$ $\left(\Delta G^{\ominus}=-421 \mathrm{~kJ}\right)$ is thermodynamically feasible.
Statement II : The above reaction occurs at room temperature.
The correct answer is
What is the enthalpy change (in J ) for converting 98 of $\mathrm{H}_2 \mathrm{O}(t)+10^{\circ} \mathrm{C}$ to $\mathrm{H}_2 \mathrm{O}(l)$ at $+20^{\circ} \mathrm{C}$ ?
$ \left(C_p\left(\mathrm{H}_2 \mathrm{O}(\eta)\right)=75 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\right) $
(density of $\mathrm{H}_2 \mathrm{O}(l)=1 \mathrm{gmL}^{-1}{ }^{})$
750
75
37.5
375
$A, B, C$ and $D$ are some compounds. The entnalpy of formation of $A(g), B(g), C(g)$ and $D(g)$ is $9.7,-110,81$ and $-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. What is $\Delta_r H$
(in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for the given reaction ?
$ A(g)+3 B(g) \longrightarrow C(g)+3 D(g) $
Observe the following reactions.
$ \begin{array}{ll} A B(g)+25 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow\left(25 \mathrm{H}_2 \mathrm{O}\right) A B ; & \Delta H=x \mathrm{~kJ} \mathrm{~mol}^{-1} \\ A B(g)+50 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow\left(50 \mathrm{H}_2 \mathrm{O}\right) A B ; & \Delta H=y \mathrm{~kJ} \mathrm{~mol}^{-1} \end{array} $
Observe the following reaction,
$ 2 A_2(g)+B_2(g) \xrightarrow{T(\mathrm{~K})} 2 A_2 B(g)+600 \mathrm{~kJ} $
The standard enthalpy of formation $\left(\Delta_f H^{\ominus}\right)$ of $A_2 B(g)$ is
Given below are two statements :
Statement I For isothermal irreversible change of an ideal gas, $q=-w=p_{\text {ext }}\left(V_{\text {final }}-V_{\text {initial }}\right)$
Statement II For adiabatic change, $\Delta U=w_{\text {adiabatic }}$
The correct answer is :
Identify the incorrect statements form the following.
I. $ \Delta S_{\text {pum }}=\left(\Delta S_{\text {nal }}+\Delta S_{\text {um }}\right) $
II. $A(\bar{i} \rightarrow A(\phi)$ : For this process entropy change decreases.
III. Entropy units are $\mathrm{JK} \mathrm{mol}^{-1}$.
What happens when methane undergoes combustion in systems A and B respectively?
| System A | System B |
|---|---|
| Temperature falls | Temperature rises |
| System A | System B |
|---|---|
| Temperature rises | Temperature remains same |
| System A | System B |
|---|---|
| Temperature falls | Temperature remains same |
| System A | System B |
|---|---|
| Temperature remains same | Temperature rises |
Given
| (A) | $\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)}$ | $\mathrm{\Delta H_1^0=-x~kJ~mol^{-1}}$ | |
|---|---|---|---|
| (B) | $\mathrm{C(graphite)+O_2(g)\to CO_2(g)}$ | $\mathrm{\Delta H_2^0=-y~kJ~mol^{-1}}$ |
$\mathrm{\Delta H^0}$ for the reaction
$\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}$ is :
Which of the following relations are correct?
(A) $\mathrm{\Delta U=q+p\Delta V}$
(B) $\mathrm{\Delta G=\Delta H-T\Delta S}$
(C) $\Delta \mathrm{S}=\frac{q_{rev}}{T}$
(D) $\mathrm{\Delta H=\Delta U-\Delta nRT}$
Choose the most appropriate answer from the options given below :
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.
[Use : $\ln 2=0.69$
Given : $S_\beta-S_\alpha=0$ at $0 \mathrm{~K}$ ]
Explanation:
$ \begin{aligned} & \mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\alpha)} \ln \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}+\mathrm{C}_{\mathrm{P}(\beta)} \ell \mathrm{n} \frac{600}{300} \\\\ & \mathrm{~S}_{\beta(600)}^{\mathrm{o}}-\mathrm{S}_{\alpha(600)}^{\mathrm{o}}=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+\left(\mathrm{C}_{\mathrm{P}(\beta)}-\mathrm{C}_{\mathrm{P}(\alpha)}\right) \ell \mathrm{n} 2 \\\\ & 6-5=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+1 \times \ell \mathrm{n} 2 \\\\ & 1=\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}+0.69 \end{aligned} $
So $\mathrm{S}_{\beta(300)}^{\mathrm{o}}-\mathrm{S}_{\alpha(300)}^{\mathrm{o}}=0.31$
Explanation:
So at $600 \mathrm{~K} \quad \Delta \mathrm{G}_{\mathrm{rxn}}^{\circ}=0$ So $\Delta \mathrm{H}^{\circ}{ }_{\text {reaction (600) }}=\mathrm{T} \Delta \mathrm{S}^{\circ}{ }_{\text {reaction (600) }}$
$\Delta \mathrm{H}^{\circ}{ }_{(600)}=600 \times 1=600 \mathrm{Joule} / \mathrm{mole}$
So $\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=\Delta \mathrm{C}_{\mathrm{P}}\left(\mathrm{T}_2-\mathrm{T}_1\right)$
$\Delta \mathrm{H}_{600}-\Delta \mathrm{H}_{300}=1 \times 300$
$\Delta \mathrm{H}_{300}=\Delta \mathrm{H}_{600}-300=600-300=300$ Joule $/$ mole.
$\mathrm{A} \rightarrow \mathrm{B}$ is an adiabatic process. If the total heat absorbed in the entire process $(\mathrm{A} \rightarrow \mathrm{B}$ and $\mathrm{B} \rightarrow \mathrm{C})$ is $\mathrm{R} T_2 \ln 10$, the value of $2 \log V_3$ is _______.
[Use, molar heat capacity of the gas at constant pressure, $C_{\mathrm{p}, \mathrm{m}}=\frac{5}{2} \mathrm{R}$ ]
Explanation:
$ \begin{aligned} &\Rightarrow \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right) =\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{V}-1} \\\\ &\Rightarrow\mathrm{~T}_1 \mathrm{~V}_1^{\gamma-1} =\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\\\ &\Rightarrow600(10)^{2 / 3} =60\left(\mathrm{~V}_2\right)^{2 / 3} \\\\ &\Rightarrow\left(\mathrm{~V}_2\right)^{2 / 3} =(10)^{5 / 3} \\\\ &\Rightarrow\mathrm{~V}_2 =(10)^{5 / 2} \\\\ &\Rightarrow\mathrm{Q}_{\mathrm{AB}} =0 \end{aligned} $
$\begin{aligned} \mathrm{Q}_{\mathrm{AC}} & =\mathrm{nRT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ & =\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) .........(i)\end{aligned}$
$\begin{gathered}\text { Total heat absorbed }=\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) \\\\ =\mathrm{RT}_2 \ln (10) ........(ii)\end{gathered}$
Equating equation (i) and equation (ii)
$ \begin{aligned} &\mathrm{RT}_2 \ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\mathrm{RT}_2 \ln (10) \\\\ &\Rightarrow\ln \left(\frac{\mathrm{V}_3}{\mathrm{~V}_2}\right) =\ln (10) \end{aligned} $
$ \mathrm{V}_3=10 \mathrm{~V}_2 $
Substitute value of $\mathrm{V}_2$
$ \begin{aligned} & =10(10)^{5 / 2}=(10)^{7 / 2} \\\\ V_3 & =(10)^{7 / 2} \end{aligned} $
Taking $\log$ on both side, we get
$ \begin{aligned} &\Rightarrow \log \left(\mathrm{V}_3\right) =\log (10)^{7 / 2} \\\\ &\Rightarrow \log \left(\mathrm{V}_3\right) =\frac{7}{2} \log (10) \\\\ &\Rightarrow 2 \log \left(\mathrm{V}_3\right) =7 \log (10) \\\\ \therefore 2 \log \left(\mathrm{V}_3\right) =7 \end{aligned} $
If $\mathrm{T}_1=2 \mathrm{~T}_2$ and $\left(\Delta \mathrm{G}_2^{\Theta}-\Delta \mathrm{G}_1^{\Theta}\right)=\mathrm{RT}_2 \ln \mathrm{x}$, then the value of $\mathrm{x}$ is _______.
$\left[\Delta \mathrm{G}_1^{\Theta}\right.$ and $\Delta \mathrm{G}_2^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $\mathrm{T}_1$ and $\mathrm{T}_2$, respectively.]
Explanation:
$\Rightarrow$ At $\mathrm{T}_1 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_1}=\frac{4}{2}=2$
$ \begin{array}{ll} \text { At } \mathrm{T}_2 \mathrm{~K}: & \mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{P}(\mathrm{g}) \\\\ \mathrm{t}=0 & 6 \\\\ \mathrm{t}=\infty & 6-\mathrm{y} \quad \mathrm{y}=2 \text { (from plot) } \end{array} $
$ \Rightarrow \text { At } \mathrm{T}_2 \mathrm{~K}: \mathrm{K}_{\mathrm{P}_2}=\frac{2}{4}=\frac{1}{2} $
Since
$ \begin{aligned} \Delta \mathrm{G}_1^0 & =-\mathrm{RT}_1 \ln \mathrm{KT}_1 ............(i) \\\\ \Delta \mathrm{G}_2^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2 ............(ii) \end{aligned} $
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =-\mathrm{RT}_2 \ln \mathrm{KT}_2+\mathrm{RT}_1 \ln \mathrm{KT}_1 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{RT}_1 \ln 2 \\\\ & =-\mathrm{RT}_2 \ln \frac{1}{2}+\mathrm{R}\left(2 \mathrm{~T}_2\right) \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2 \ln 2+2 \mathrm{RT}_2 \ln 2\end{aligned}$
$\begin{aligned} \Delta \mathrm{G}_2^0-\Delta \mathrm{G}_1^0 & =3 \mathrm{RT}_2 \ln 2 \\\\ & =\mathrm{RT}_2\left(\ln 2^3\right) \\\\ & =\mathrm{RT}_2 \ln 8\end{aligned}$
At $T(\mathrm{~K}) 2$ mole of an ideal gas is allowed to expand reversibly and isothermally from a pressure of 10 atmospheres to 1 atmosphere. The work done (in kJ ) is $\left(R=8.3 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$
$-3.82 \times 10^{-1} \times T$
$-4.82 \times 10^{-1} \times T$
$-2.82 \times 10^{-2} \times T$
$-3.82 \times 10^{-2} \times T$
For the reaction at $25^{\circ} \mathrm{C}, X_2 \mathrm{O}_4(l) \longrightarrow 2 \mathrm{XO}_2(g), \Delta U$ and $\Delta S$ are 2.1 k cal and $20 \mathrm{cal} / \mathrm{K}$ respectively. What is $\Delta \mathrm{G}$ for the reaction at the same temperature? ( $R=2 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )
-2.67 k cal
+2.67 k cal
-1.67 k cal
+3.67 k cal