Question

Change in enthalpy for reaction
2H2O2(l) → 2H2O(l) + O2(g)
If heat of formation of H2O2(l) and H2O(l) are
– 188 & – 286 KJ/mol respectively : -
(1) – 196 KJ/mol (2) + 196 KJ/mol
(3) + 948 KJ/mol (4) – 948 KJ/mo

Answer 1

1)-196KJ/mol
heat of for=2×-286KJ/mol-2×-188KJ/mol
= -196

Answer 2

Answer : The correct option is, (1) -196 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2H_2O_2(l)\rightleftharpoons +2H_2O(l)+O_2(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[n_{(H_2O)}\times \Delta H^o_f_{(H_2O)}+n_{(O_2)}\times \Delta H^o_f_{(O_2)}]-[n_{(H_2O_2)}\times \Delta H^o_f_{(H_2O_2)}]$

We are given:

$\Delta H^o_f_{(H_2O_2(g))}=-188kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-286kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2mol\times -286kJ/mol)+(1mol\times 0kJ/mol)]-[(2mol\times -188kJ/mol)]=-196kJ/mol$

Therefore, the change in enthalpy of the reaction is -196 kJ/mol