The heat of neutralization of $HCl$ and $NaOH$ is:

  • A
    $0 \ kJ \ mol^{-1}$
  • B
    $-57.3 \ kJ \ mol^{-1}$
  • C
    $+57.3 \ kJ \ mol^{-1}$
  • D
    None of these

Explore More

Similar Questions

Determine the enthalpy of formation for $H_2O_2(\ell)$,using the listed enthalpies of reaction:
$N_2H_{4(\ell)} + 2H_2O_{2(\ell)} \to N_{2(g)} + 4H_2O_{(\ell)}; \Delta _r H_1^o = -818 \, kJ/mol$
$N_2H_{4(\ell)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(\ell)}; \Delta _r H_2^o = -622 \, kJ/mol$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(\ell)}; \Delta _r H_3^o = -285 \, kJ/mol$
Calculate the value in $kJ/mol$.

The heat of transition $(\Delta H_t)$ of graphite into diamond would be,where
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = x \ kJ \ mol^{-1}$
$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = y \ kJ \ mol^{-1}$

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......$kJ/mol$

The enthalpy change $(\Delta H)$ for the neutralisation of $1 \ M \ HCl$ by caustic potash in dilute solution at $298 \ K$ is ..... $kJ$.

If at $298 \, K$ the bond energies of $C-H, C-C, C=C$ and $H-H$ bonds are respectively $414, 347, 615$ and $435 \, kJ \, mol^{-1}$,the value of enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ will be $.... \, kJ$.

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