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
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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
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heat of for=2×-286KJ/mol-2×-188KJ/mol
= -196
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
The equation used to calculate enthalpy change is of a reaction is:
The equilibrium reaction follows:
The equation for the enthalpy change of the above reaction is:
We are given:
Putting values in above equation, we get:
Therefore, the change in enthalpy of the reaction is -196 kJ/mol