The enthalpy change for the reaction H2O2(l)⟶H2O(l)+(1/2)O2(g) is −98.3 kJ mol−1 If the enthalpy of formation of H2O2(l) is −187.4 kJ mol−1the enthalpy of formation of H2O(l) would be

# The enthalpy change for the reaction ${\mathrm{H}}_{2}{\mathrm{O}}_{2}\left(\mathrm{l}\right)⟶{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)+\left(1/2\right){\mathrm{O}}_{2}\left(\mathrm{g}\right)$ is  If the enthalpy of formation of ${\mathrm{H}}_{2}{\mathrm{O}}_{2}\left(\mathrm{l}\right)$ is the enthalpy of formation of ${\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)$ would be

1. A

2. B

3. C

4. D

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### Solution:

The reaction to be considered is ${\mathrm{H}}_{2}\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_{2}\left(\mathrm{g}\right)⟶{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)$

Given data are

$\begin{array}{l}{\mathrm{H}}_{2}{\mathrm{O}}_{2}\left(\mathrm{l}\right)⟶{\mathrm{H}}_{2}\mathrm{O}\left(\mathrm{l}\right)+\left(1/2\right){\mathrm{O}}_{2}\left(\mathrm{g}\right) \mathrm{\Delta }H=-98.3\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{H}}_{2}\left(\mathrm{g}\right)+{\mathrm{O}}_{2}\left(\mathrm{g}\right)⟶{\mathrm{H}}_{2}{\mathrm{O}}_{2}\left(\mathrm{l}\right) \mathrm{\Delta }H=-187.4\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Adding these two reactions gives the required reaction. Hence  