JEE MAIN - Chemistry (2023 - 1st February Morning Shift - No. 16)
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
499
Explanation
$\frac{\text { (i) }+\text { (iii) }}{2}-$ (ii) gives desired reaction
$$ \begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\frac{436+78}{2}-(-242) \\\\ & =\frac{436+78}{2}+242=499 \end{aligned} $$
$$ \begin{aligned} & \Delta \mathrm{H}_{\mathrm{r}}=\frac{436+78}{2}-(-242) \\\\ & =\frac{436+78}{2}+242=499 \end{aligned} $$
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