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प्रश्न
The enthalpy of formation of nitrogen dioxide is +33.2 kJ mol−1. The enthalpy of the reaction \[\ce{2N2_{(g)} + 4O2_{(g)} -> 4NO2_{(g)}}\]; is ____________.
पर्याय
−33.2 kJ
+33.2 kJ
−66.4 kJ
+132.8 kJ
MCQ
रिकाम्या जागा भरा
उत्तर
The enthalpy of formation of nitrogen dioxide is +33.2 kJ mol−1. The enthalpy of the reaction \[\ce{2N2_{(g)} + 4O2_{(g)} -> 4NO2_{(g)}}\]; is +132.8 kJ.
Explanation:
∆fH refers to the formation of one mole of NO2.
i.e. \[\ce{1/2 N2_{(g)} + O2_{(g)} -> NO2_{(g)}}\]; ∆fH = +33.2 kJ
∴ ∆H for 4 moles = 4 mol × 33.2 kJ = +132.8 kJ.
∴ Enthalpy of the given reaction is +132.8 kJ
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Thermochemistry
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