Question

Heat of formation of $H_{2}O$ is -188 kJ/mol and $H_2{O}_2$ is 286 kJ/mol. The enthalpy change for the reaction is:
$2H_2{O}_2\to 2H_{2}O+O_2$

• A. 196 kJ
• B. -196 kJ
• C. 984 kJ
• D. -984 kJ

Answer 1

Answer: (D)

-984 kJ

Heat of formation of H2O is -188 kJ/mol
$H_2+0.5O_2\to H_{2}O$; $\Delta H_1=-188$ kJ/mol ....(1)
Heat of formation of H2O2 is 286 kJ/mol
$H_2+O_2\to H_2{O}_2$; $\Delta H_2=286$ kJ/mol ....(2)
Multiply reaction (1) with 2.
$2H_2+O_2\to 2H_{2}O$; $\Delta H_3=-376$ kJ/mol ....(3)
Multiply reaction (2) with 2.
$2H_2+2O_2\to 2H_2{O}_2$; $\Delta H_4=572$ kJ/mol ....(4)
(3) - (4)
Subtract equation (4) from the equation (3)
$2H_2{O}_2\to 2H_{2}O+O_2$
The enthalpy change for the reaction is $\Delta H_3-\Delta H_4=-376-572=-948$ kJ/mol