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NCERT Class XI Chemistry Thermodynamics Solutions

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Question : 8 of 22
Marks: +1, -0
The reaction of cyanamide, NH2CN(s)\mathrm{NH_2CN_{(s)}} , with dioxygen was carried out in a bomb calorimeter, and DU was found to be –742.7 kJ mol−1\mathrm{kJ\,mol^{-1}} at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH2CN(s)+32O2(g)\mathrm{NH_2CN_{(s)}} + \frac{3}{2}\mathrm{O_{2(g)}} → N2(g)+CO2(g)+H2O(l)\mathrm{N_{2(g)}} + \mathrm{CO_{2(g)}} + \mathrm{H_2O_{(l)}}
Solution:  
NH2CN(s)+32O2(g)\mathrm{NH_2CN_{(s)}} + \frac{3}{2}\mathrm{O_{2(g)}} → N2(g)+CO2(g)+H2O(l)\mathrm{N_{2(g)}} + \mathrm{CO_{2(g)}} + \mathrm{H_2O_{(l)}}
ΔU = - 742.7 kJ/mol
Δn = np−nRn_p - n_R = 2 - 32\frac{3}{2} = + 12\frac{1}{2} , R = 8.3141000\frac{8.314}{1000} kJ K−1 mol−1\mathrm{kJ\,K^{-1}\,mol^{-1}} ; T = 298 K
ΔH = ΔU + ΔnRT = - 742.7 + 12×8.3141000\frac{1}{2} \times \frac{8.314}{1000} × 298
= –742.7 + 1.2 = – 741.5 kJ/mol
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