Question

Change in enthalpy for reaction, $2H_2{O}_2\left(l\right)$ $\to$ $2H_{2}O\left(l\right)$ are -188 and -286 kj/mol respectively, is

• A. -196 kj/mol
• B. +196 kj/mol
• C. +948 kj/mol
• D. -948 kj/mol

Answer 1

The correct option is A -196 kj/mol

$\begin{array}{c}\Delta H=\left\{2\Delta H\left(H_{2}o\right)+\Delta H\left(O_2\right)\right\}-2\Delta \left(H_2{O}_2\right)\\ \Delta H\left(H_{2}O\right)=-286kJ/mol\\ \Delta H\left(O_{2}g\right)=0\\ \Delta H\left(H_2{O}_2\right)=-188kJ/mol\\ \Delta H=2\left(-286\right)+0-2\times \left(188\right)\\ =-196kJ/mol\end{array}$

Hence,

Option $A$ is correct answer.