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Question

The enthalpy of formation of ammonia is $$-46.0\ kJ\ mol^{-1}$$. The enthalpy change for the reaction $$2NH_3(g)\rightarrow 2N_2(g)+3H_2(g)$$ is:


A
46.0 kJ mol1
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B
92.0 kJ mol1
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C
23.0 kJ mol1
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D
92.0 kJ mol1
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Solution

The correct option is B $$92.0\ kJ\ mol^{-1}$$
1968611_1026452_ans_5845deb10071457fbf1643403939758d.jpg

Chemistry

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