Question

Change in enthalpy for reaction
$2H_2O_2\left(l\right)\to 2H_2O\left(l\right) + O_2\left(g\right)$
if heat of formation of $H_2O_2\left(l\right)$ and $H_2O\left(l\right)$ 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

Answer: (A)

$2H_{2}O_{2}\left(l\right)\to2H_{2}O\left(l\right)+O_{2} \Delta H=?$
$\Delta H=\left[\left(2\times\Delta H_{f} of H_{2}O\left(l\right)+\Delta H_{f} of O_{2} \right)\right]-\left[\left(2\times\Delta H_{f} of H_{2}O_{2}(l)\right)\right]$
$\left[\left(2\times-286\right)+\left(0\right)-\left(2\times188\right)\right]$
$=\left[-572+376\right]=-196\,kJ/mol$