On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.

  • A
    $-393$
  • B
    $-655$
  • C
    $393$
  • D
    $655$

Explore More

Similar Questions

The atomization enthalpies of $NH_{3(g)}$ and $N_2H_{4(g)}$ are $+150 \ kJ \ mol^{-1}$ and $+310 \ kJ \ mol^{-1}$ respectively. The $\Delta H(N-N)$ bond enthalpy in $kJ \ mol^{-1}$ is:

The heat of formation of methane $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ at constant pressure is $-18500 \ cal$ at $25 \ ^oC$. The heat of reaction at constant volume would be (in $cal$)

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:

Calculate the enthalpy change for the following reaction, using the given bond energies $(\text{kJ/mol})$: $C-H = 414$, $H-O = 463$, $H-Cl = 431$, $C-Cl = 326$, and $C-O = 335$.
$CH_3OH(g) + HCl(g) \rightarrow CH_3Cl(g) + H_2O(g)$

The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$

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