For the reaction,$H_{2(g)} + I_{2(g)} \to 2HI_{(g)}$,$\Delta H = -12.40 \ kcal$. According to this,the heat of formation of $HI$ will be......$kcal$.

  • A
    $12.4$
  • B
    $-12.4$
  • C
    $-6.20$
  • D
    $6.20$

Explore More

Similar Questions

For the reaction $C(s) + O_2(g) \rightarrow CO_2(g)$,the enthalpy change $(\Delta H)$ is:

Given that bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively and $\Delta H_f$ for $HCl$ is $-90 \ kJ \ mol^{-1},$ the bond enthalpy of $HCl$ is ............... $kJ \ mol^{-1}$.

If for the reaction $CCl_{4(g)} \rightarrow C_{(g)} + 4Cl_{(g)}$ the following data is given:
$\Delta_{vap} H^{\theta} (CCl_{4(l)}) = 30 \ kJ \ mol^{-1}$
$\Delta_{f} H^{\theta} (CCl_{4(l)}) = -136.0 \ kJ \ mol^{-1}$
$\Delta_{a} H^{\theta} (C_{(s)}) = 714.0 \ kJ \ mol^{-1}$
$\Delta_{a} H^{\theta} (Cl_{2(g)}) = 242.0 \ kJ \ mol^{-1}$
Calculate the mean bond enthalpy of $C-Cl$ in $CCl_{4(g)}$.

$H_{2(g)} + Cl_{2(g)} \to 2HCl_{(g)}, \Delta H = -44 \ kcal$
$2Na_{(s)} + 2HCl_{(g)} \to 2NaCl_{(s)} + H_{2(g)}, \Delta H = -152 \ kcal$
For the reaction $Na_{(s)} + \frac{1}{2}Cl_{2(g)} \to NaCl_{(s)}, \Delta H = \dots \ kcal$

The standard enthalpy of formation of $NH_3$ is $-46.0 \, kJ/mol$. If the enthalpy of formation of $H_2$ from its atoms is $-436 \, kJ/mol$ and that of $N_2$ is $-712 \, kJ/mol$,the average bond enthalpy of $N-H$ bond in $NH_3$ is......$kJ/mol$

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