$C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$; $\Delta H = -3267.7 \ kJ \ mol^{-1}$. Given that the standard enthalpies of formation of $CO_{2(g)}$ and $H_2O_{(g)}$ are $-393.5 \ kJ \ mol^{-1}$ and $-285.85 \ kJ \ mol^{-1}$ respectively,calculate the standard enthalpy of formation of benzene $(C_6H_{6(l)})$.
(A) The combustion reaction is: $C_6H_{6(l)} + 7.5O_{2(g)} \to 6CO_{2(g)} + 3H_2O_{(g)}$
$\Delta H_{comb} = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + 7.5 \times \Delta H_f(O_2)]$
Given $\Delta H_{comb} = -3267.7 \ kJ \ mol^{-1}$,$\Delta H_f(CO_2) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f(H_2O) = -285.85 \ kJ \ mol^{-1}$,and $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$.
$-3267.7 = [6 \times (-393.5) + 3 \times (-285.85)] - \Delta H_f(C_6H_6)$
$-3267.7 = [-2361 - 857.55] - \Delta H_f(C_6H_6)$
$-3267.7 = -3218.55 - \Delta H_f(C_6H_6)$
$\Delta H_f(C_6H_6) = -3218.55 + 3267.7 = 49.15 \ kJ \ mol^{-1}$.
$\Delta H_{comb} = [6 \times \Delta H_f(CO_2) + 3 \times \Delta H_f(H_2O)] - [\Delta H_f(C_6H_6) + 7.5 \times \Delta H_f(O_2)]$
Given $\Delta H_{comb} = -3267.7 \ kJ \ mol^{-1}$,$\Delta H_f(CO_2) = -393.5 \ kJ \ mol^{-1}$,$\Delta H_f(H_2O) = -285.85 \ kJ \ mol^{-1}$,and $\Delta H_f(O_2) = 0 \ kJ \ mol^{-1}$.
$-3267.7 = [6 \times (-393.5) + 3 \times (-285.85)] - \Delta H_f(C_6H_6)$
$-3267.7 = [-2361 - 857.55] - \Delta H_f(C_6H_6)$
$-3267.7 = -3218.55 - \Delta H_f(C_6H_6)$
$\Delta H_f(C_6H_6) = -3218.55 + 3267.7 = 49.15 \ kJ \ mol^{-1}$.
Explore More
Similar Questions
The bond enthalpies of heavy hydrogen $(D-D)$, oxygen $(O=O)$, and heavy water $(D-O)$ are $+400$, $+498$, and $+490 \ kJ \ mol^{-1}$, respectively. The $\Delta_{r} H^{\circ}$ of the reaction to produce $D_2O$ is:
From the following,the heat of neutralization is maximum in:
Easy
View SolutionThe standard enthalpy of formation of $CO_{(g)}$, $CO_{2(g)}$, $N_2O_{(g)}$ and $N_2O_{4(g)}$ are respectively $-110$, $-393$, $81$ and $-10 \ kJ \ mol^{-1}$. The enthalpy change (in $kJ$) of the following reaction is $N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$
Calculate the standard enthalpy of formation of $ICl_{(g)}$ based on the following reactions. The standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$ respectively.
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:
Difficult
View SolutionDefine thermochemical equations and explain the conventions used in them.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo