Question

If $\Delta H_f^o$ for $H_2{O}_2\left(l\right)$ and $H_{2}O\left(l\right)$ are $-188kJ$ ${mol}^{-1}$ and $-286kJ$ ${mol}^{-1}$, what will be the enthalpy change of the reaction?
$2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$

• A. $146kJ$ ${mol}^{-1}$
• B. $-196kJ$ ${mol}^{-1}$
• C. $-494kJ$ ${mol}^{-1}$
• D. $-98kJ$ ${mol}^{-1}$

Answer 1

Answer: (B)

$-196kJ$ ${mol}^{-1}$

For the given reaction,
$2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$
enthalpy change of the reaction = $\Delta H$ = $2\times \Delta H_{f\left(H_{2}O\right)}-2\times \Delta H_{f\left(H_2{O}_2\right)}$ (since, $\Delta H_{f\left(O_2\right)}=0$ )
hence, $\Delta H=2\times (-286)-2\times (-188)=-196KJ{\text{mol}}^{-1}$
Hence, answer is option $B$.