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Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ?rH for the reaction: N2O4(g) + 3CO(g) ? N2O(g) + 3CO2(g) ?

sudhanshu , 11 Years ago
Grade 12
anser 2 Answers
Priyanshu kumar

Last Activity: 8 Years ago

∆rH=[sum of H(product)]-[sum of H(reactant)]∆rH=[1×H(n2o)+3×H(co2)]-[1×H(n2o4)+3×H(co)]=[1×81+3×-393]-[1×9.7+3×-110]=[81-1179]-[9.7-330]=(-1098)-(-320.3)=(-1098+320.3)∆rH=(-777.7)KJ
T ranjith kumar

Last Activity: 7 Years ago

Heat of combustion of ph3 gas is 228 kilocalories more inverse and that to be 284 gas is 355 kilocalories more than was the energy of the p2p bond is in kilo calories
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