${1 \over 2} X_2$ + ${3 \over 2} Y_2 \to$ XY3, $\Delta H$ = -30 kJ, to be at equilibrium, the temperature will be :
${1 \over 2}C{l_2}(g)$ $\buildrel {{1 \over 2}{\Delta _{diss}}{H^\Theta }} \over \longrightarrow $ $Cl(g)$ $\buildrel {{\Delta _{eg}}{H^\Theta }} \over \longrightarrow $ $C{l^ - }(g)$ $\buildrel {{\Delta _{Hyd}}{H^\Theta }} \over \longrightarrow $ $C{l^ - }(aq)$
(Using the data, ${\Delta _{diss}}H_{C{l_2}}^\Theta $ = 240 kJ/mol, ${\Delta _{eg}}H_{Cl}^\Theta $ = -349 kJ/mol, ${\Delta _{hyd}}H_{C{l^ - }}^\Theta $ = - 381 kJ/mol) will be :
and $R = 8.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$ )
CaCO3(s) $\to$ CaO(s) + CO2 (g) the vales of ∆H° and ∆S° are +179.1 kJ mol−1 and 160.2 J/K respectively at 298 K and 1 bar. Assuming that ∆H° do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is :
Cl2(g) = 2Cl(g), 242.3 kJ mol–1
I2(g) = 2I(g), 151.0 kJ mol–1
ICl(g) = I(g) + Cl(g), 211.3 kJ mol–1
I2(s) = I2(g), 62.76 kJ mol–1
Given that the standard states for iodine and chlorine are I2(s) and Cl2(g), the standard enthalpy of formation for ICl(g) is :
H2C = CH2(g) + H2(g) $\to$ H3C - CH3(g) at 298 K will be :
2C + O2 $\to$ 2CO2; $\Delta H$ = -393 J
2Zn + O2 $\to$ 2ZnO; $\Delta H$ = -412 J
If the enthalpy of sublimation of Li is $155 \mathrm{~kJ} \mathrm{~mol}^{-1}$, enthalpy of dissociation of $\mathrm{F}_2$ is $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$, ionization enthalpy of Li is $520 \mathrm{~kJ} \mathrm{~mol}^{-1}$, electron gain enthalpy of F is $-313 \mathrm{~kJ} \mathrm{~mol}^{-1}$, standard enthalpy of formation of LiF is $-594 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The magnitude of lattice enthalpy of LiF is $\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)
Explanation:
$ \begin{aligned} &\text { Use the following data : }\\ &\begin{array}{|c|c|c|} \hline \text { Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJ} \mathrm{~mol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{~g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{~g}) & x & 280 \\ \hline \end{array} \end{aligned} $
One mole each of $\mathrm{A}_2(\mathrm{~g})$ and $\mathrm{B}_2(\mathrm{~g})$ are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K .
$ \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) $
The value of $x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$ . (Nearest integer)
(Given : $\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )
Explanation:
For the reaction at $500\ \text{K}$:
$\mathrm{A_2(g)}+\mathrm{B_2(g)} \rightleftharpoons 2\mathrm{AB(g)}$
1) Calculate $\Delta H^\circ$ of reaction at $500\ \text{K}$
Using standard enthalpies of formation:
$ \Delta H^\circ = \sum \nu \Delta_f H^\circ(\text{products})-\sum \nu \Delta_f H^\circ(\text{reactants}) $
$ \Delta H^\circ = 2(32) - \big(6 + x\big) = 64 - 6 - x = (58-x)\ \text{kJ mol}^{-1} $
2) Calculate $\Delta S^\circ$ of reaction at $500\ \text{K}$
$ \Delta S^\circ = 2(222) - (146+280) = 444 - 426 = 18\ \text{J K}^{-1}\text{mol}^{-1} $
3) Calculate $\Delta G^\circ$ using $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$
Convert $T\Delta S^\circ$ into kJ:
$ T\Delta S^\circ = 500 \times 18 = 9000\ \text{J mol}^{-1} = 9\ \text{kJ mol}^{-1} $
So,
$ \Delta G^\circ = (58-x) - 9 = (49-x)\ \text{kJ mol}^{-1} $
4) Use $\Delta G^\circ=-RT\ln K$
Given $\log K = 2.2$ (base 10),
$ \ln K = 2.2\ln 10 = 2.2 \times 2.303 = 5.0666 $
Now,
$ \Delta G^\circ = -RT\ln K = -(8.3)(500)(5.0666)\ \text{J mol}^{-1} $
$ \Delta G^\circ = -21026\ \text{J mol}^{-1} \approx -21.0\ \text{kJ mol}^{-1} $
5) Equate and solve for $x$
$ 49 - x = -21 $
$ x = 70\ \text{kJ mol}^{-1} $
Nearest integer: $\boxed{70}$
Resonance in $\mathrm{X}_2 \mathrm{Y}$ can be represented as
The enthalpy of formation of $X_2Y$ $ \left(X = X(g) + \frac{1}{2} Y = Y(g) \rightarrow X_2Y(g) \right) $ is 80 kJ mol$^{-1}$. The magnitude of resonance energy of $X_2Y$ is __ kJ mol$^{-1}$ (nearest integer value).
Given: Bond energies of $X \equiv X$, $X = X$, $Y = Y$ and $X = Y$ are 940, 410, 500, and 602 kJ mol$^{-1}$ respectively.
valence $X$: 3, $Y$: 2
Explanation:
To calculate the magnitude of the resonance energy ($\Delta \mathrm{H}_{\text{R.E.}}$) of the compound $X_2Y$, follow these steps:
Understand the formula: The resonance energy is given by the equation:
$ \Delta \mathrm{H}_{\text{R.E.}} = \Delta \mathrm{H}_{\mathrm{f(exp)}} - \Delta \mathrm{H}_{\mathrm{f(Theo)}} $
Given values:
Experimental enthalpy of formation for $X_2Y(g)$, $\Delta \mathrm{H}_{\mathrm{f(exp)}} = 80 \text{ kJ/mol}$.
Bond energies:
$ \mathrm{BE}_{X \equiv X} = 940 \text{ kJ/mol} $
$ \mathrm{BE}_{X = X} = 410 \text{ kJ/mol} $
$ \mathrm{BE}_{Y = Y} = 500 \text{ kJ/mol} $
$ \mathrm{BE}_{X = Y} = 602 \text{ kJ/mol} $
Calculate theoretical enthalpy of formation ($\Delta \mathrm{H}_{\mathrm{f(Theo)}}$):
The reaction considered:
$ \mathrm{X}_{2(g)} + \frac{1}{2} \mathrm{Y}_{2(g)} \rightarrow \mathrm{X}_2 \mathrm{Y}_{(g)} $
The energy involved is calculated as:
$ \Delta \mathrm{H}_{\mathrm{f(Theo)}} = \left(\mathrm{BE}_{X = X} + \frac{1}{2} \mathrm{BE}_{Y = Y}\right) - \left(\mathrm{BE}_{X = X} + \mathrm{BE}_{X = Y}\right) $
Plugging in the values:
$ = \left(940 + \frac{1}{2} \times 500\right) - (410 + 602) $
$ = 1190 - 1012 = 178 \text{ kJ/mol} $
Calculate resonance energy:
Using the resonance energy formula:
$ \Delta \mathrm{H}_{\text{R.E.}} = 80 - 178 = -98 \text{ kJ/mol} $
The magnitude (absolute value) of the resonance energy is:
$ |\Delta \mathrm{H}_{\text{R.E.}}| = 98 \text{ kJ/mol} $
Therefore, the magnitude of the resonance energy of $X_2Y$ is $\boxed{98}$ kJ/mol.
A perfect gas ( 0.1 mol ) having $\overline{\mathrm{C}}_v=1.50 \mathrm{R}$ (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is $(-)$___________$J$ (nearest integer)
[Given: $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]
Explanation:
To determine the total work done when a perfect gas undergoes a transformation from point 1 to point 4, we can analyze each step of the process:
Given Data
Moles of gas, $n = 0.1 \, \text{mol}$
Specific heat at constant volume, $\overline{\mathrm{C}}_v=1.50 \mathrm{R}$
Gas constant, $\mathrm{R}=0.082 \, \mathrm{L} \, \mathrm{atm} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$
Work Done Calculation
Step 1: From Point 1 to Point 2 ($W_{1 \rightarrow 2}$)
Since this is an isochoric process (constant volume), the work done, $W_{1 \rightarrow 2}$, is zero.
$ \mathrm{W}_{1 \rightarrow 2} = 0 $
Step 2: From Point 2 to Point 3 ($W_{2 \rightarrow 3}$)
This step is isobaric (constant pressure), and the work done can be calculated as:
$ \mathrm{W}_{2 \rightarrow 3} = -P\Delta V = -P(V_3 - V_2) $
Given:
$P = 3 \, \text{atm}$
Change in volume, $\Delta V = V_3 - V_2 = 2 - 1 = 1 \, \text{L}$
$ \mathrm{W}_{2 \rightarrow 3} = -3 \, \text{atm} \times 1 \, \text{L} = -3 \, \text{L atm} $
Converting to Joules (using $1 \, \text{L atm} = 101.3 \, \text{J}$):
$ \mathrm{W}_{2 \rightarrow 3} = -3 \times 101.3 \, \text{J} = -304 \, \text{J} $
Step 3: From Point 3 to Point 4 ($W_{3 \rightarrow 4}$)
This is another isochoric process, so the work done, $W_{3 \rightarrow 4}$, is zero.
$ \mathrm{W}_{3 \rightarrow 4} = 0 $
Total Work Done
The total work done from point 1 to point 4 is the sum of work done in each step:
$ \text{Total work done} = W_{1 \rightarrow 2} + W_{2 \rightarrow 3} + W_{3 \rightarrow 4} = 0 + (-304) + 0 = -304 \, \text{J} $
Thus, the total work done during this transformation is $-304 \, \text{J}$.
Explanation:
Calculate the moles of octane:
$ \text{Moles of octane} = \frac{1.14 \, \text{g}}{114 \, \text{g/mol}} = 0.01 \, \text{moles} $
Determine the heat evolved during combustion:
The heat capacity of the calorimeter is $ 5 \, \text{kJ/K} $ and the temperature increase is $ 5 \, \text{K} $.
$ \text{Heat evolved} = C \times \Delta T = 5 \, \text{kJ/K} \times 5 \, \text{K} = 25 \, \text{kJ} $
Calculate the magnitude of the heat of combustion:
The heat of combustion per mole of octane is found by dividing the total heat evolved by the moles of octane combusted:
$ \text{Magnitude of Heat of combustion} = \frac{25 \, \text{kJ}}{0.01 \, \text{moles}} = 2500 \, \text{kJ/mol} $
Therefore, the magnitude of the heat of combustion of octane at constant volume is $ 2500 \, \text{kJ/mol} $.
Given :
$ \begin{aligned} & \left.\Delta \mathrm{H}^{\ominus}{ }_{\text {sub }}[\mathrm{C} \text { (graphite })\right]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}}=\mathrm{C} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $
The $\Delta \mathrm{H}_{\mathrm{f}} \ominus$ for $\mathrm{CH}_2=\mathrm{CH}_2$ is_________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)
Explanation:
$\begin{aligned} & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{fl})}}=2 \times\left[\Delta \mathrm{H}_{\mathrm{sub}}^{\mathrm{o}}\right]_{\mathrm{C}_{(\mathrm{s})}}+2 \times \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}^{\mathrm{o}}-1 \times \Delta \mathrm{H}_{\mathrm{C}=\mathrm{C}}^{\mathrm{o}}-4 \times \Delta \mathrm{H}_{\mathrm{C}-\mathrm{H}}^{\circ} } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=(2 \times 710)+(2 \times 436)-611-4 \times 414 } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=25 \mathrm{~kJ} / \mathrm{mol} } \end{aligned}$
Consider the following data :
Heat of formation of $\mathrm{CO}_2(\mathrm{g})=-393.5 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Heat of formation of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286.0 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$
Heat of combustion of benzene $=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
The heat of formation of benzene is __________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
Explanation:
To determine the heat of formation of benzene ($ \Delta H_f[\text{C}_6 \text{H}_6] $), we use the given data:
Heat of formation of $\text{CO}_2(\text{g}) = -393.5 \text{ kJ/mol}$
Heat of formation of $\text{H}_2 \text{O}(\text{l}) = -286.0 \text{ kJ/mol}$
Heat of combustion of benzene = $-3267.0 \text{ kJ/mol}$
The reaction for the combustion of benzene is:
$ \text{C}_6\text{H}_6 + \frac{15}{2} \text{O}_2(\text{g}) \rightarrow 6 \text{CO}_2(\text{g}) + 3 \text{H}_2\text{O}(\text{l}) $
Using the formula for the enthalpy change of the reaction ($ \Delta H_R $):
$ \Delta H_R = \Delta H_C = \Sigma \Delta H_f(\text{Products}) - \Sigma \Delta H_f(\text{Reactants}) $
Substitute values into the equation:
$ -3267 = 6 \times (-393.5) + 3 \times (-286) - \Delta H_f[\text{C}_6\text{H}_6] $
Solving this equation, we find:
$ \Delta H_f[\text{C}_6\text{H}_6] = 48 \text{ kJ/mol} $
Thus, the heat of formation of benzene is $ 48 \text{ kJ/mol} $.
The formation enthalpies, $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{H}_{(\mathrm{g})}$ and $\mathrm{O}_{(\mathrm{g})}$ are 220.0 and $250.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively, at 298.15 K , and $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ is $-242.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at the same temperature. The average bond enthalpy of the $\mathrm{O}-\mathrm{H}$ bond in water at 298.15 K is _______ $\mathrm{kJ} \mathrm{~mol}^{-1}$ (nearest integer).
Explanation:
$\begin{array}{ll} \frac{1}{2} \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{H}(\mathrm{~g}) \quad ;\Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{H}_{(\mathrm{g})}\right)=220 \mathrm{KJ} / \mathrm{mol} \\ \frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{O}(\mathrm{~g}) \quad ; \Delta_{\mathrm{f}} \mathrm{H}\left(\mathrm{O}_{(\mathrm{g})}\right)=250 \mathrm{KJ} / \mathrm{mol} \end{array}$
$\begin{aligned} & \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}\right)=-242=440+250-2(\mathrm{~B} . \mathrm{E} .(\mathrm{O}-\mathrm{H})) \\ & \mathrm{BE}(\mathrm{O}-\mathrm{H})=466 \mathrm{KJ} / \mathrm{mol} \end{aligned}$
Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_5$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction
$\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$
will be at equilibrium is __________ (Nearest integer)
Explanation:
To determine the temperature at which the given reaction is at equilibrium, we need to apply the concept of Gibbs free energy change $(\Delta G^0)$ at equilibrium, where $\Delta G^0$ is equal to zero.
The given reaction is:
$ \frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 $
We are given the standard entropies and enthalpy change ($\Delta H^{\ominus}$):
Standard entropies:
$\mathrm{X}_2 = 70 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $
$\mathrm{Y}_2 = 50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $
$\mathrm{XY}_5 = 110 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $
Standard enthalpy change:
$\Delta H^{\ominus} = -35 \, \mathrm{kJ/mol}$
First, calculate the standard entropy change ($\Delta S_{Rxn}^0$) of the reaction:
$ \Delta S_{Rxn}^0 = 110 - \left(\frac{1}{2} \times 70 + \frac{5}{2} \times 50\right) $
$ = 110 - \left(35 + 125\right) $
$ = 110 - 160 $
$ = -50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1} $
At equilibrium, $\Delta G^0 = 0$, and the relation $\Delta G^0 = \Delta H^0 - T\Delta S^0$ applies. Thus:
$ 0 = -35000 \, \mathrm{J/mol} - T(-50 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}) $
Solving for $T$:
$ T \cdot 50 = 35000 $
$ T = \frac{35000}{50} $
$ T = 700 \, \mathrm{K} $
Therefore, the temperature at which the reaction is at equilibrium is 700 Kelvin.
The bond dissociation enthalpy of $\mathrm{X}_2 \Delta \mathrm{H}_{\text {bond }}^{\circ}$ calculated from the given data is ___________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)
$\begin{aligned} & \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\text {lattice }}^{\circ}=800 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{M}(\mathrm{~s}) \rightarrow \mathrm{M}(\mathrm{~g}) \Delta \mathrm{H}_{\text {sub }}^{\circ}=100 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$
$\mathrm{M}(\mathrm{~g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{i}}^{\circ}=500 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{X}(\mathrm{~g})+\mathrm{e}^{-}(\mathrm{g}) \rightarrow \mathrm{X}^{-}(\mathrm{g}) \Delta \mathrm{H}_{\mathrm{eg}}^{\circ}=-300 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\mathrm{M}(\mathrm{~s})+\frac{1}{2} \mathrm{X}_2(\mathrm{~g}) \rightarrow \mathrm{M}^{+} \mathrm{X}^{-}(\mathrm{s}) \Delta \mathrm{H}_f^{\circ}=-400 \mathrm{~kJ} \mathrm{~mol}^{-1}$
[Given : $\mathrm{M}^{+} \mathrm{X}^{-}$is a pure ionic compound and X forms a diatomic molecule $\mathrm{X}_2$ in gaseous state]
Explanation:
$\begin{aligned} & \begin{aligned} \therefore \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{MX}) & =\Delta \mathrm{H}_{\text {sub }}(\mathrm{M})+\text { I.E. }(\mathrm{M})+\frac{1}{2}[\text { B.E. }(\mathrm{X}-\mathrm{X})] \\ & +\mathrm{EG}(\mathrm{X})+\text { L.E. }(\mathrm{MX}) \end{aligned} \\ & -400=(100)+(500)+\frac{1}{2}(\text { B.E. })+(-300)+(-800)\\ & \therefore \text { B.E. }=200 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned} ~ \begin{aligned} \end{aligned}$
The standard enthalpy and standard entropy of decomposition of $\mathrm{N}_2 \mathrm{O}_4$ to $\mathrm{NO}_2$ are $55.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $175.0 \mathrm{~J} / \mathrm{K} / \mathrm{mol}$ respectively. The standard free energy change for this reaction at $25^{\circ} \mathrm{C}$ in J $\mathrm{mol}^{-1}$ is ________ (Nearest integer)
Explanation:
$\Delta \mathrm{H}_{\mathrm{rxn}}^{\circ}=55 \mathrm{~kJ} / \mathrm{mol}, \quad \mathrm{T}=298 \mathrm{~K}$
$\begin{aligned} & \Delta \mathrm{S}_{\mathrm{rxn}}^{\mathrm{o}}=175 \mathrm{~J} / \mathrm{mol} \\ & \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=\Delta \mathrm{H}_{\mathrm{rxn}}^{\mathrm{o}}-\mathrm{T} \Delta \mathrm{~S}_{\mathrm{rxn}}^{\mathrm{o}} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000 \mathrm{~J} / \mathrm{mol}-298 \times 175 \mathrm{~J} / \mathrm{mol} \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=55000-52150 \\ & \Rightarrow \Delta \mathrm{G}_{\mathrm{rxn}}^{\mathrm{o}}=2850 \mathrm{~J} / \mathrm{mol} \end{aligned}$
Consider the following cases of standard enthalpy of reaction $\left(\Delta \mathrm{H}_{\mathrm{r}}^{\circ}\right.$ in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$
$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_1^{\circ}=-1550 \\ & \mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta \mathrm{H}_2^{\circ}=-393.5 \\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \Delta \mathrm{H}_3^{\circ}=-286 \end{aligned}$
The magnitude of $\Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})}^{\circ}$ is ____________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (Nearest integer).
Explanation:
$ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+\frac{7}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_1^{\circ} =-1550 \ldots \text { (i) } \\\\ & \mathrm{C} \text { (graphite) }+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \Delta \mathrm{H}_2^{\circ}=-393.5 \ldots \text { (ii) } \\\\ & \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \Delta \mathrm{H}_3^{\circ}=-286 .......(iii) \end{aligned} $
$\begin{aligned} & \text { From } 2 \times \mathrm{eq}^{\mathrm{n}}(\mathrm{ii})+3 \times \mathrm{eq}^{\mathrm{n}}(\mathrm{iii})-\mathrm{eq}^{\mathrm{n}}(\mathrm{i}) \\\\ & 2 \mathrm{C}_{\text {(graphite) }}+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}): \Delta \mathrm{H}_{f \mathrm{C}_2 \mathrm{H}_6}^{\circ} \\\\ & \left(\Delta \mathrm{H}_f^{\circ}\right)_{\mathrm{C}_2 \mathrm{H}_6}=2 \times(-393.5)+3 \times(-286)-(-1550) \\\\ & =-95 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$When $\Delta \mathrm{H}_{\mathrm{vap}}=30 \mathrm{~kJ} / \mathrm{mol}$ and $\Delta \mathrm{S}_{\mathrm{vap}}=75 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, then the temperature of vapour, at one atmosphere is _________ K.
Explanation:
To find the temperature of vaporization at one atmosphere, we can use the Clausius-Clapeyron equation, which relates the enthalpy of vaporization ($\Delta H_{vap}$) to the change in entropy ($\Delta S_{vap}$) during the phase transition at a particular temperature (T). The relationship can be simplified under the assumption that both the enthalpy of vaporization and the entropy change of vaporization are constant with temperature to the form:
$\Delta H_{vap} = T \cdot \Delta S_{vap}$
This equation states that the enthalpy change of vaporization is equal to the product of the temperature at which the phase change occurs and the entropy change of vaporization. We rearrange this equation to solve for the temperature (T):
$T = \frac{\Delta H_{vap}}{\Delta S_{vap}}$
However, it's crucial to ensure the units are consistent. Given that $\Delta H_{vap}$ is in kJ/mol and $\Delta S_{vap}$ is in J/(mol·K), we need to convert $\Delta H_{vap}$ from kJ/mol to J/mol to match units:
$\Delta H_{vap} = 30 \, \text{kJ/mol} = 30 \times 10^3 \, \text{J/mol}$
Substituting the given values into the equation, we obtain:
$T = \frac{30 \times 10^3 \, \text{J/mol}}{75 \, \text{J/(mol·K)}}$
$T = \frac{30000 \, \text{J/mol}}{75 \, \text{J/(mol·K)}}$
$T = 400 \, \text{K}$
Therefore, the temperature of vaporization at one atmosphere is 400 K.
When equal volume of $1 \mathrm{~M} \mathrm{~HCl}$ and $1 \mathrm{~M} \mathrm{~H}_2 \mathrm{SO}_4$ are separately neutralised by excess volume of $1 \mathrm{M}$ $\mathrm{NaOH}$ solution. $x$ and $y \mathrm{~kJ}$ of heat is liberated respectively. The value of $y / x$ is __________.
Explanation:
To solve this problem, we need to understand the concept of neutralization reactions and the heat evolved during these reactions.
When an acid and a base react, they undergo a neutralization reaction to produce water and a salt. The heat released in this process is known as the enthalpy of neutralization.
Consider the neutralization of hydrochloric acid (HCl) and sulfuric acid (H2SO4) by sodium hydroxide (NaOH). The balanced chemical equations for these neutralizations are:
$ \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} $
$ \mathrm{H_2SO_4} + 2\mathrm{NaOH} \rightarrow \mathrm{Na_2SO_4} + 2\mathrm{H_2O} $
For the first reaction, each mole of HCl reacts with one mole of NaOH, releasing a certain amount of heat (let's denote this amount by $ x $ kJ). For the second reaction, each mole of H2SO4 reacts with two moles of NaOH. Since we are given equal volumes and molarities of HCl and H2SO4, we can infer that one mole of H2SO4 will produce twice the heat of one mole of HCl because it produces double the amount of water.
Thus, the heat evolved in the neutralization of H2SO4 by NaOH (denoted as $ y $ kJ) will be approximately twice that of HCl. Therefore, $ y = 2x $.
Therefore, the value of $\frac{y}{x}$ is:
$ \frac{y}{x} = \frac{2x}{x} = 2 $
So, the value of $\frac{y}{x}$ is 2.
The heat of solution of anhydrous $\mathrm{CuSO}_4$ and $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ are $-70 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $+12 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
The heat of hydration of $\mathrm{CuSO}_4$ to $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is $-x \mathrm{~kJ}$. The value of $x$ is ________. (nearest integer).
Explanation:
(I) $\mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$ Solution $\Delta \mathrm{H}=-70 \mathrm{~kJ}$
(II) $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$ Solution $\mathrm{\Delta H=12 \mathrm{~kJ}}$
(I) - (II)
$\begin{aligned} & \mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \\ & \Delta \mathrm{H}=-70-12=-82 \end{aligned}$
$\Delta_{\text {vap }} \mathrm{H}^{\ominus}$ for water is $+40.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at 1 bar and $100^{\circ} \mathrm{C}$. Change in internal energy for this vapourisation under same condition is ________ $\mathrm{kJ} \mathrm{~mol}^{-1}$. (Integer answer) (Given $\mathrm{R}=8.3 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
To find the change in internal energy for the vaporization of water under the given conditions, we'll use the following relationship between enthalpy change ($\Delta_{\text{vap}} H$) and internal energy change ($\Delta_{\text{vap}} U$):
$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \Delta V$
For vaporization, the change in volume ($\Delta V$) can be approximated by considering the volume of the vapor because the volume of liquid water is relatively small compared to the volume of the vapor.
The ideal gas law gives us:
$P V = n R T$
Since we are dealing with 1 mole of water:
$V = \frac{R T}{P}$
Plugging this into the enthalpy change equation, we get:
$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + P \left( \frac{R T}{P} \right)$
This simplifies to:
$\Delta_{\text{vap}} H = \Delta_{\text{vap}} U + R T$
Rearranging for $\Delta_{\text{vap}} U$:
$\Delta_{\text{vap}} U = \Delta_{\text{vap}} H - R T$
Given:
$\Delta_{\text{vap}} H = 40.79 \, \text{kJ mol}^{-1}$ (or 40790 J/mol)
$R = 8.3 \, \text{JK}^{-1} \text{mol}^{-1}$
$T = 100^\circ \text{C} + 273.15 = 373.15 \, \text{K}$
Now, substitute the values into the equation:
$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - (8.3 \, \text{JK}^{-1} \text{mol}^{-1} \times 373.15 \, \text{K})$
$\Delta_{\text{vap}} U = 40790 \, \text{J mol}^{-1} - 3097.145 \, \text{J mol}^{-1}$
$\Delta_{\text{vap}} U = 37692.855 \, \text{J mol}^{-1}$
Converting back to kJ:
$\Delta_{\text{vap}} U = 37.69 \, \text{kJ mol}^{-1}$
Rounding to the nearest integer, the change in internal energy for the vaporization of water under the given conditions is:
38 kJ mol-1
Consider the figure provided.
$1 \mathrm{~mol}$ of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at $18^{\circ} \mathrm{C}$. If the piston is moved to position $\mathrm{B}$, keeping the temperature unchanged, then '$\mathrm{x}$' $\mathrm{L}$ atm work is done in this reversible process.
$\mathrm{x}=$ ________ $\mathrm{L}$ atm. (nearest integer)
[Given : Absolute temperature $={ }^{\circ} \mathrm{C}+273.15, \mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}{ }^{-1} \mathrm{~K}^{-1}$]
Explanation:
$\begin{aligned} & \mathrm{V}_1=100 \mathrm{~L} \\ & \mathrm{~V}_2=10 \mathrm{~L} \\ & \mathrm{~W}=-\mathrm{nR} \operatorname{Tl} \frac{\mathrm{V}_2}{\mathrm{~V}_1} \\ & =-1 \times 0.08206 \times 291.15 \times 2.303 \log \frac{10}{100} \\ & =55 \mathrm{~L} \text { atm } \\ & \end{aligned}$
For the reaction at $298 \mathrm{~K}, 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}, \Delta \mathrm{H}=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta S=0.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. The reaction will become spontaneous above __________ $\mathrm{K}$.
Explanation:
To determine the temperature above which the reaction $2A+B \rightarrow C$ becomes spontaneous, we can use the Gibbs free energy equation:
$\Delta G = \Delta H - T\Delta S$
The reaction becomes spontaneous when $\Delta G$ is negative. Therefore, we need to find the temperature at which $\Delta G$ changes from positive to negative. We set $\Delta G$ to zero to find the threshold temperature:
$0 = \Delta H - T\Delta S$
Substituting the given values of $\Delta H = 400 \, \text{kJ mol}^{-1} = 400,000 \, \text{J mol}^{-1}$ (since 1 kJ = 1000 J) and $\Delta S = 0.2 \, \text{kJ mol}^{-1} K^{-1} = 200 \, \text{J mol}^{-1} K^{-1}$, we get:
$0 = 400,000 \, \text{J mol}^{-1} - T(200 \, \text{J mol}^{-1} K^{-1})$
Solving for $T$, we have:
$T = \frac{400,000 \, \text{J mol}^{-1}}{200 \, \text{J mol}^{-1} K^{-1}}$
$T = 2000 \, \text{K}$
Therefore, the reaction will become spontaneous above $2000 \, \text{K}$. This means that at temperatures higher than 2000 K, the reaction tends towards product formation without the need for external energy to drive the process.
An ideal gas, $\overline{\mathrm{C}}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}$, is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is $298 \mathrm{~K}$ and $5 \mathrm{~atm}$, respectively then the final temperature is _________ $\mathrm{K}$ (nearest integer).
[$\overline{\mathrm{c}}_{\mathrm{v}}$ is the molar heat capacity at constant volume]
Explanation:
$-1\left(2 V_1-V_1\right)=n \times \frac{5 R}{2}\left(T_2-T_1\right)$
$\begin{aligned} & -\mathrm{V}_1=\frac{5}{2}\left(n R T_2-5 \mathrm{~V}_1\right) \\ & -\mathrm{V}_1=2.5\left(\mathrm{nRT_{2 } )}-12.5 \mathrm{~V}_1\right. \\ & 11.5 \mathrm{~V}_1=2.5\left(\mathrm{nRT_{2 } )}\right. \\ & 11.5 \times \frac{n R T_1}{P_1}=2.5 \times\left(n R T_2\right) \\ & \mathrm{T}_2=274.16 \mathrm{~k} \\ & \approx 274 \text { (Nearest integer) } \\ & \end{aligned}$
Combustion of 1 mole of benzene is expressed at
$\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \text {. }$
The standard enthalpy of combustion of $2 \mathrm{~mol}$ of benzene is $-^{\prime} x^{\prime} \mathrm{kJ}$. $x=$ __________.
Given :
1. standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$, for the reaction $6 \mathrm{C}$ (graphite) $+3 \mathrm{H}_2(\mathrm{g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})$ is $48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
2. Standard Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{CO}_2(\mathrm{g})$, for the reaction $\mathrm{C}$ (graphite) $+\mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})$ is $-393.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
3. Standard and Enthalpy of formation of $1 \mathrm{~mol}$ of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$, for the reaction $\mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})$ is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Explanation:
To determine the standard enthalpy of combustion of 2 moles of benzene, we need to use the standard enthalpy of formation values provided and apply Hess's Law. Here is a step-by-step explanation:
Given Data:
- Standard enthalpy of formation of benzene ($C_6H_6(l)$):
$ \Delta H_f(\text{C}_6\text{H}_6(l)) = 48.5 \, \text{kJ/mol} $
- Standard enthalpy of formation of carbon dioxide ($CO_2(g)$):
$ \Delta H_f(\text{CO}_2(g)) = -393.5 \, \text{kJ/mol} $
- Standard enthalpy of formation of water ($H_2O(l)$):
$ \Delta H_f(\text{H}_2O(l)) = -286 \, \text{kJ/mol} $
Combustion Reaction for Benzene:
$ \text{C}_6\text{H}_6(l) + \frac{15}{2} \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 3 \text{H}_2O(l) $
Enthalpy Change Calculation:
Using Hess's Law, the enthalpy change for the reaction can be calculated as follows:
$ \Delta H_{\text{comb}} = \left[ 6 \Delta H_f(\text{CO}_2(g)) + 3 \Delta H_f(\text{H}_2O(l)) \right] - \Delta H_f(\text{C}_6\text{H}_6(l)) $
Substitute the given values:
$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) + 3 \times (-286) \right] - 48.5 $
Perform the calculations:
$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) \right] + \left[ 3 \times (-286) \right] - 48.5 $
$ \Delta H_{\text{comb}} = \left[ -2361 \right] + \left[ -858 \right] - 48.5 $
$ \Delta H_{\text{comb}} = -3267.5 \, \text{kJ/mol} $
This value is the enthalpy change for the combustion of 1 mole of benzene.
For 2 Moles of Benzene:
$ \Delta H_{\text{comb (2 moles)}} = 2 \times (-3267.5 \, \text{kJ/mol}) $
$ \Delta H_{\text{comb (2 moles)}} = -6535 \, \text{kJ} $
Conclusion:
The standard enthalpy of combustion of 2 moles of benzene is
$ x = 6535 \, \text{kJ} $
Thus, $ x = 6535 $.
The heat of combustion of solid benzoic acid at constant volume is $-321.30 \mathrm{~kJ}$ at $27^{\circ} \mathrm{C}$. The heat of combustion at constant pressure is $(-321.30-x \mathrm{R}) \mathrm{~kJ}$, the value of $x$ is __________.
Explanation:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}(\mathrm{s})+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 7 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{I})$
$\begin{aligned} & \Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{n}_9 R T \\ & \Delta \mathrm{H}=-321.30-\frac{1}{2} \mathrm{R} \times 300 \end{aligned}$
So, $x=150$
Three moles of an ideal gas are compressed isothermally from $60 \mathrm{~L}$ to $20 \mathrm{~L}$ using constant pressure of $5 \mathrm{~atm}$. Heat exchange $\mathrm{Q}$ for the compression is - _________ Lit. atm.
Explanation:
For isothermal process
$\Rightarrow \begin{aligned} & Q=-W \\ & Q=-5 \times 40 \\ & |Q|=+200 \text { Lit atm } \end{aligned}$
The enthalpy of formation of ethane $(\mathrm{C}_2 \mathrm{H}_6)$ from ethylene by addition of hydrogen where the bond-energies of $\mathrm{C}-\mathrm{H}, \mathrm{C}-\mathrm{C}, \mathrm{C}=\mathrm{C}, \mathrm{H}-\mathrm{H}$ are $414 \mathrm{~kJ}, 347 \mathrm{~kJ}, 615 \mathrm{~kJ}$ and $435 \mathrm{~kJ}$ respectively is $-$ __________ $\mathrm{kJ}$
Explanation:
$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 \longrightarrow \mathrm{C}_2 \mathrm{H}_6 \\ & \begin{aligned} \Delta \mathrm{H} & =(615)+(435)-(347)-2(414) \\ & =615+435-347-828 \\ & =-125 \mathrm{~kJ} \end{aligned} \end{aligned}$
(Given $\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ )
Explanation:
To determine the change in standard Gibbs free energy ($\Delta G^{\circ}$) for the reaction at a given temperature when the equilibrium constant ($K$) is known, we can use the following relationship:
$ \Delta G^{\circ} = -RT \ln K $
Here, $R$ is the universal gas constant ($8.314 \ J K^{-1} mol^{-1}$), $T$ is the temperature in Kelvin ($300 \ K$), and $K$ is the equilibrium constant.
Puting the values we get,
$ \Delta G^{\circ} = -8.314 \times 300 \times \ln(10) $
We can convert the natural logarithm of 10 to base-10 logarithm, using the change of base formula, $\ln(10) = 2.3026$. So,
$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 $
Calculating this yields:
$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 = -5743.914 \ J mol^{-1} $
To convert this to kilojoules per mole, we divide by 1000:
$ \Delta G^{\circ} = -5.743914 \times 10^{3} \ J mol^{-1} \times \frac{1 \ kJ}{10^{3} J} = -5.743914 \ kJ mol^{-1} $
Now, when expressing this in terms of $\times 10^{-1} \ kJ mol^{-1}$, we get:
$ \Delta G^{\circ} = -57.43914 \times 10^{-1} \ kJ mol^{-1} $
Therefore, $\Delta G^{\circ}$ for the reaction is approximately $-57.43914 \times 10^{-1} \ kJ mol^{-1}$ or $-57.4 \times 10^{-1} \ kJ mol^{-1}$ when rounded to three significant figures.
If 5 moles of an ideal gas expands from $10 \mathrm{~L}$ to a volume of $100 \mathrm{~L}$ at $300 \mathrm{~K}$ under isothermal and reversible condition then work, $\mathrm{w}$, is $-x \mathrm{~J}$. The value of $x$ is __________.
(Given R = 8.314 J K$^{-1}$ mol$^{-1}$)
Explanation:
It is isothermal reversible expansion, so work done negative
$\begin{aligned} & \mathrm{W}=-2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right) \\ & =-2.303 \times 5 \times 8.314 \times 300 \log \left(\frac{100}{10}\right) \\ & =-28720.713 \mathrm{~J} \\ & \equiv-28721 \mathrm{~J} \end{aligned}$
Consider the following reaction at $298 \mathrm{~K} \cdot \frac{3}{2} \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{O}_{3(g)} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29}$. $\Delta_r G^{\ominus}$ for the reaction is _________ $\mathrm{kJ}$. (Given $\mathrm{R}=8.314 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}$)
Explanation:
$\begin{aligned} & \frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{O}_{3(\mathrm{~g})} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29} . \\ & \Delta_{\mathrm{r}} \mathrm{G}^{\Theta}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{P}} \\ & =-8.314 \times 10^{-3} \times 298 \times \ln \left(2.47 \times 10^{-29}\right) \\ & =-8.314 \times 10^{-3} \times 298 \times(-65.87) \\ & =163.19 \mathrm{~kJ} \end{aligned}$
Two reactions are given below:
$\begin{aligned} & 2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol} \\ & \mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$
Then enthalpy change for following reaction $3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}$ is _______ $\mathrm{kJ} / \mathrm{mol}$.
Explanation:
$2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\circ}=-822 \mathrm{~kJ} / \mathrm{mol}$ ........ (1)
$\mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\circ}=-110 \mathrm{~kJ} / \mathrm{mol}$ ........ (2)
$3 \mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}_3=\text { ? }$
$\begin{aligned} & (3)=3 \times(2)-(1) \\ & \begin{aligned} \Delta \mathrm{H}_3 & =3 \times \Delta \mathrm{H}_2-\Delta \mathrm{H}_1 \\ & =3(-110)+822 \\ & =492 \mathrm{~kJ} / \mathrm{mole} \end{aligned} \end{aligned}$