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Question

The standard enthalpy of formation of N2O and NO are 28 kJ mol1 and 90 kJ mol1 respectively. The enthalpy change of the reaction, 2N2O(g)+O2(g)4NO(g) is equal to:

A
62 kJ mol1
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B
88 kJ mol1
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C
252 kJ mol1
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D
304 kJ mol1
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Solution

The correct option is D 304 kJ mol1
First write the balanced chemical equation for the formation N2O and NO:

N2+12O2N2O ;ΔHo=28 kJ mol1 ...(i)

12N2+12O2NO ;ΔHo=90 kJ mol1 ...(ii)

Then multiply equation (i) by 2 and (ii) equation by 4 we get,
2N2+O22N2O ;ΔHo=2×28 kJ mol1 ....(iii)

2N2+2O24NO ;ΔHo=4×90 kJ mol1....(iv)

Now substract (iii) equation from (iv) equation, we get:
2N2O(g)+O2(g)4NO(g) ;ΔHo=304 kJ mol1
Thus, the enthalpy change of the reaction is 304 kJ mol1.

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