Calculate ?H rxn for the following reaction. 2H 2 O 2 (l) = 2H 2 O(l) + O 2 (g)

Question

Calculate ?Hrxn for the following reaction.

2H2O2(l) = 2H2O(l) + O2(g)

given that ?Hac[H2O(l)] = -285.8 kJ/mol and ?Hac[H2O2(l)] = -87.6 kJ/mol.

53.0 kJ

98.2 kJ

?98.2 kJ

196.4 kJ

?396.4 kJ

Please explain why. I have found people say both -396.4 and 196.4.

Calculate ?Hrxn for the following reaction.

2H2O2(l) = 2H2O(l) + O2(g)

given that ?Hac[H2O(l)] = -285.8 kJ/mol and ?Hac[H2O2(l)] = -87.6 kJ/mol.

53.0 kJ

98.2 kJ

?98.2 kJ

196.4 kJ

?396.4 kJ

Please explain why. I have found people say both -396.4 and 196.4.

Explanation / Answer

Given,

?Hf[H2O]=-285.8 kJ/mol

?Hf[H2O2]=-87.6 kJ/mol

?Hf[O2]=0 kJ/mol

For the reaction:

?Hreaction=stoichimetric coeff of product*[?Hf(product)]- stoichimetric coeff of reactant* [?Hf(reactant)]

=2*(-285.8)-[2*(-87.6)]

=-396.4 kJ

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