Q. 83.7( 59 Votes )

# The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and DU was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

We know that,

ΔU = ΔH - ΔngRT

The balanced chemical equation of reaction is,

NH2CN(s) + 3/2O2(g) N2(g) + CO2(g) + H2O(l)

Number of moles of gaseous particles in product side is 2 (one N2 and one CO2)

Number of moles of gaseous particles on reactant side is 3/2 (O2)

Thus, change in number of moles of gaseous particle,

Δng = 2 – 3/2 = 1/2

ΔU = ΔH - ΔngRT

Given,

ΔU = -742.7 kJ

R =8.314 J K-1 mol-1

= 8.414 × 10-3 kJ K-1 mol-1

Δng =(2-2.5) =-0.5 mole

T = 298 K

Thus,

ΔH = (-742.7 kJ mol-1) – (1/2mol)×(8.314 × 10-3 kJ K-1mol-1)×298

= -742.7 - 1.239

= -743.9 kJ

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