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If \Delta H_{f}^{o} for H_2O_2(l) and H_2O(l) are -188 kJ/mole and -286 kJ/mol respectively. What will be the enthalpy change of the given reaction (in kJ/mole) ?

2 H _{2} O _{2}(l) \rightarrow 2 H _{2} O ( l )+ O _{2}( g )

Option: 1

-196


Option: 2

146


Option: 3

-494


Option: 4

-98


Answers (1)

best_answer

For the given reaction:

\mathrm{2 H _{2} O _{2}(l) \rightarrow 2 H _{2} O ( l )+ O _{2}( g )}

The enthalpy change for the reaction is defined as

\mathrm{\Delta H_R = 2\times H_{f,\ H_2O}^0+ H_{f,\ O_2}^0-2 \times H_{f,\ H_2O_2}^0}

\therefore \mathrm{\Delta H_R = 2\times (-286)+ 0 -2 \times (-188)= -196}

Hence, the correct answer is Option(1)

Posted by

Ritika Harsh

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