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Question

The value of ∆H (formation) for NH3 is -91.8 kJ per mol than calculate enthalpy change for 2NH3(g)-----> N2(g) + 3H2(g)


A
18.36KJ
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B
1836KJ
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C
183.6KJ
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D
1.836KJ
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Solution

The correct option is C 183.6KJ
ΔH for formation is given. For the reverse reaction,ΔHchanges sign as the reverse of exothermic reaction will be endothermic. So, ΔH for decomposition is - (-91.8)=91.8 for one mole. But here, two moles are decomposing,
ΔH=291.8=183.6KJ

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