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

  • A
    $4.519$
  • B
    $45.10$
  • C
    $451.9$
  • D
    $45.19$

Explore More

Similar Questions

The heat of neutralization of a strong dibasic acid by a dilute solution of $NaOH$ is approximately ....... $Kcal/equivalent$.

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 enthalpy of reaction for the reaction: $2 H_{2(g)} + O_{2(g)} \to 2 H_{2}O_{(l)}$ is $\Delta_{r}H^{\theta} = -572 \ kJ \ mol^{-1}$. What will be the standard enthalpy of formation of $H_{2}O_{(l)}$?

Given that,$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H^{\circ} = -x \ kJ \ mol^{-1}$ and $2CO_{(g)} + O_{2(g)} \longrightarrow 2CO_{2(g)} ; \Delta H^{\circ} = -y \ kJ \ mol^{-1}$. The enthalpy of formation of $CO$ will be:

Match the following allotropes of carbon with their standard enthalpy of formation $(\Delta_f H^{\Theta})$:
Allotrope$\Delta_f H^{\Theta}$
$i$. Graphite$b$. $0 \ kJ/mol$
$ii$. Diamond$c$. $1.90 \ kJ/mol$
$iii$. Fullerene$a$. $38.1 \ kJ/mol$

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