At 25^{circ} C, the enthalpy of the following | vedclass

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(A) To find the enthalpy change for the reaction $H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)}$, we use Hess's Law by manipulating the given equations:$1

Vedclass · Accepted Answer

$499$

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(A) To find the enthalpy change for the reaction $H_2O_{(g)} ightarrow H_{(g)} + OH_{(g)}$, we use Hess's Law by manipulating the given equations:$1$. $2 H_2O_{(g)} ightarrow 2 H_{2(g)} + O_{2(g)} \quad \Delta H^{\circ} = -2 	imes (-242) = +484 \ kJ \ mol^{-1}$$2$. $H_{2(g)} + O_{2(g)} ightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = +78 \ kJ \ mol^{-1}$$3$. $H_{2(g)} ightarrow 2 H_{(g)} \quad \Delta H^{\circ} = +436 \ kJ \ mol^{-1}$Adding these equations:$2 H_2O_{(g)} + H_{2(g)} + O_{2(g)} + H_{2(g)} ightarrow 2 H_{2(g)} + O_{2(g)} + 2 OH_{(g)} + 2 H_{(g)}$Simplifying, we get: $2 H_2O_{(g)} ightarrow 2 H_{(g)} + 2 OH_{(g)}$$\Delta H^{\circ} = 484 + 78 + 436 = +998 \ kJ \ mol^{-1}$For the reaction $H_2O_{(g)} ightarrow H_{(g)} + OH_{(g)}$, the enthalpy change is:$X = \frac{998}{2} = 499 \ kJ \ mol^{-1}$

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