The standard enthalpy of formation of N2O and NO are 28 kJ mol−1 and 90 kJ mol−1 respectively. The enthalpy change of the reaction, 2N2O(g)+O2(g)→4NO(g) is equal to:
A
62kJ mol−1
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B
88 kJ mol−1
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C
−252 kJ mol−1
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D
304 kJ mol−1
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Solution
The correct option is D304 kJ mol−1 First write the balanced chemical equation for the formation N2O and NO:
N2+12O2→N2O;ΔHo=28 kJ mol−1...(i)
12N2+12O2→NO;ΔHo=90 kJ mol−1...(ii)
Then multiply equation (i) by 2 and (ii) equation by 4 we get, 2N2+O2→2N2O;ΔHo=2×28 kJ mol−1....(iii)
2N2+2O2→4NO;ΔHo=4×90 kJ mol−1....(iv)
Now substract (iii) equation from (iv) equation, we get: 2N2O(g)+O2(g)→4NO(g);ΔHo=304 kJ mol−1
Thus, the enthalpy change of the reaction is 304 kJ mol−1.