Question

Change in enthalpy for the reaction, $2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$, if heat of formation of $H_2{O}_2\left(l\right)$ and $H_{2}O\left(l\right)$ are $-188$ and $-286kJ/mol$ respectively is:

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

Answer 1

Answer: (A)

$-196kJ/mol$

$2H_2{O}_2\left(l\right)⟶2H_{2}O\left(l\right)+O_2\left(g\right);\Delta H=$?

$\Delta H=[(2\times \Delta H_{f}of{H}_{2}O\left(l\right))+\left(\Delta H_{f}of{O}_2\right)-(2\times \Delta H_{f}of{H}_2{O}_2\left(l\right))]$

$=[(2\times -286)+\left(0\right)-(2\times -188)]$

$=[-572+376]=-196kJ/mol$