Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

  • A
    $-22.88$
  • B
    $-228.88$
  • C
    $228.88$
  • D
    $-343.52$

Explore More

Similar Questions

Which of the following reactions defines the standard enthalpy of combustion,$\Delta H_c^ \circ$?

Calculate the enthalpy change for the reaction
$H_2 + F_2 \longrightarrow 2HF$
given that
Bond energy of $H-H$ bond $= 434 \ kJ/mol$
Bond energy of $F-F$ bond $= 158 \ kJ/mol$
Bond energy of $H-F$ bond $= 565 \ kJ/mol$
Result in $kJ$.

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

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:

Given:
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?

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