Question

6.80 g of NH3; is passed over heated CuO. The standard heat enthalpies of NH3, CuO(s) and H2O(l) are -46.0, -155.0 and −2850kJmol−1 respectively and the change is.NH3+(3/2)CuO⟶12N2(g)+(3/2)H2O(l)+(3/2)Cu(s).If the enthalpy change is

A
-50
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B
-59.6
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C
60
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D
30
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Solution

The correct option is B -59.6NH3+(3/2)CuO⟶12N2(g)+(3/2)H2O(l)+(3/2)Cu(s);ΔH=? ΔH∘=ΣH∘Products−ΣH∘Reactants =[12×H∘N2+(3/2)×H∘H2O(l)+ (3/2)H∘Cu]−[(3/2)H∘CuO+H∘NH3] =[0+(3/2)×(−285)+(3/2)×0]−[(3/2)×(−155)+(−46)] =−149kJ [∵H∘forpureelement=0andH∘compound=H∘formation]∵17gNH3bringsinchangeinΔH=149kJ∵6.8gNH3bringsinchangeinΔH=(149×6.8)/17=59.6kJ∴ΔHfor6.89gNH3=−59.6kJ

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