The enthalpy of reaction for the reaction: 2 | vedclass

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(A) The given reaction is: $2 H_{2(g)} + O_{2(g)} 	o 2 H_{2}O_{(l)}$ with $\Delta_{r}H^{	heta} = -572 \ kJ \ mol^{-1}$.The standard enthalpy of form

Vedclass · Accepted Answer

$-286 \ kJ \ mol^{-1}$

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(A) The given reaction is: $2 H_{2(g)} + O_{2(g)} 	o 2 H_{2}O_{(l)}$ with $\Delta_{r}H^{	heta} = -572 \ kJ \ mol^{-1}$.The standard enthalpy of formation $(\Delta_{f}H^{	heta})$ is defined as the enthalpy change when $1 \ mol$ of a compound is formed from its constituent elements in their standard states.The formation reaction for $H_{2}O_{(l)}$ is: $H_{2(g)} + \frac{1}{2} O_{2(g)} 	o H_{2}O_{(l)}$.This reaction is exactly half of the given reaction. Therefore,the enthalpy of formation is half of the given enthalpy of reaction:$\Delta_{f}H^{	heta}(H_{2}O) = \frac{\Delta_{r}H^{	heta}}{2} = \frac{-572 \ kJ \ mol^{-1}}{2} = -286 \ kJ \ mol^{-1}$.

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 $H_{2}O_{(l)}$?

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