Given the following thermochemical equations:
$1) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -787 \ kJ$
$2) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}, \Delta H = -286 \ kJ$
$3) \ C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}, \Delta H = -1310 \ kJ$
Calculate the enthalpy of formation of acetylene $(C_2H_{2(g)})$ in $kJ \ mol^{-1}$.

  • A
    $+1802$
  • B
    $-1802$
  • C
    $-800$
  • D
    $+237$

Explore More

Similar Questions

Consider the reaction $2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)$. The magnitude of enthalpy change for the reaction in $\text{kJ mol}^{-1}$ is . . . . . . . (Nearest integer). Given: $\Delta_f H^\circ(H_2S) = -20.1 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(H_2O) = -286.0 \text{ kJ mol}^{-1}$, $\Delta_f H^\circ(SO_2) = -297.0 \text{ kJ mol}^{-1}$

For the reaction $N_2 + 3 X_2 \longrightarrow 2 NX_3$,where $X = F, Cl$ (the average bond energies are $F-F = 155 \ kJ \ mol^{-1}$,$N-F = 272 \ kJ \ mol^{-1}$,$Cl-Cl = 242 \ kJ \ mol^{-1}$,$N-Cl = 200 \ kJ \ mol^{-1}$ and $N \equiv N = 941 \ kJ \ mol^{-1}$),the heats of formation of $NF_3$ and $NCl_3$ in $kJ \ mol^{-1}$,respectively,are closest to

If the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:

Identify the $INVALID$ equation.

Enthalpy of a compound is equal to its

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