Question

The enthalpy of reaction, $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(g\right)$ is $\Delta H_1$ and that of following reaction is $\Delta H_2$.

$H_{2_{\left(g\right)}}+\frac{1}{2}{O}_2\left(g\right)\to H_2{O}_{\left(l\right)}$

Then :

• A. $\Delta H_1<\Delta H_2$
• B. $\Delta H_1+\Delta H_2=0$
• C. $\Delta H_1>\Delta H_2$
• D. $\Delta H_1=\Delta H_2$

Answer 1

The correct option is A $\Delta H_1<\Delta H_2$

$\Delta H_1=H_{1}reactant-H_{1}product$

$\Delta H_2=H_{2}reactant-H_{2}product$

$H_{1}reactant=H_{2}reactant$ But,

$H_{1}product>H_{2}reactant$

$\Delta H_2-\Delta H_1=$ $H_{1}product-H_{2}reactant>0$

Hence, $\Delta H_2>\Delta H_1$

Hence the correct option is A.