At $25^{\circ} \mathrm{C}$, the enthalpy of the following processes are given :
| $\mathrm{H_2(g)+O_2(g)}$ | $\to$ | $2\mathrm{OH(g)}$ | $\mathrm{\Delta H^\circ=78~kJ~mol^{-1}}$ |
|---|---|---|---|
| $\mathrm{H_2(g)+\frac{1}{2}O_2(g)}$ | $\to$ | $\mathrm{H_2O(g)}$ | $\mathrm{\Delta H^\circ=-242~kJ~mol^{-1}}$ |
| $\mathrm{H_2(g)}$ | $\to$ | $\mathrm{2H(g)}$ | $\mathrm{\Delta H^\circ=436~kJ~mol^{-1}}$ |
| $\frac{1}{2}\mathrm{O_2(g)}$ | $\to$ | $\mathrm{O(g)}$ | $\mathrm{\Delta H^\circ=249~kJ~mol^{-1}}$ |
What would be the value of X for the following reaction ? _____________ (Nearest integer)
$\mathrm{H_2O(g)\to H(g)+OH(g)~\Delta H^\circ=X~kJ~mol^{-1}}$
Answer (integer)
499
Solution
$\frac{\text { (i) }+\text { (iii) }}{2}-$ (ii) gives desired reaction
<br/><br/>$$
\begin{aligned}
& \Delta \mathrm{H}_{\mathrm{r}}=\frac{436+78}{2}-(-242) \\\\
& =\frac{436+78}{2}+242=499
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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