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  • The enthalpy of reaction for the reaction: 2H_{2} (g) + O_{2}(g) \rightarrow 2H_{2}O(l) is \Delta_{R}H\ominus = - 572 kJ \; mol^{-1}. What will be standard enthalpy of formation of H_{2}O(I) ?

The enthalpy of reaction for the reaction:
2H_{2} (g) + O_{2}(g) \rightarrow 2H_{2}O(l) is \Delta_{r}H^{\circleddash }= - 572 kJ \; mol^{-1}. What will be standard enthalpy of formation of H_{2}O(l) ?

Answers (1)

Standard molar enthalpy of formation is the enthalpy change for the formation of one mole of a compound from its most stable states or reference states. As per the given information in the question, the standard enthalpy for the given equation is – 572 kJ mol–1

Now the enthalpy of formation for H_{2}O will be half the enthalpy of the value in the given equation. So, now we can calculate that:

\Delta _{f}H=\frac{1}{2}\times \Delta _{y}H=-\frac{52}{2}=-286\; kJ\; mol^{-1}

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