Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

  • A
    $91$
  • B
    $92$
  • C
    $93$
  • D
    $95$

Explore More

Similar Questions

Define thermochemical equations and explain the conventions used in them.

Difficult
View Solution

The conversion of oxygen to ozone represented by the equation $3O_2 \to 2O_3$ is an endothermic reaction. The enthalpy change $\Delta H$ accompanying the reaction:

What amount of energy (in $KJ$) is released in the combustion of $5.8\, g$ of $C_4H_{10}\,(g)$?
$2 C_4H_{10}(g) + 13 O_2(g) \to 8 CO_2(g) + 10 H_2O(l)$; $\Delta H^o = -5756\, KJ$

For a reaction $A + B + 30 \ kcal \longrightarrow \text{Product}$,the energy of activation of its backward reaction is $50 \ kcal$. The energy of activation of its forward reaction is :-

The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo