Home NCERT Solutions NCERT Exemplar CBSE Sample Papers NCERT Books Class 12 Class 10
Question 8

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

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

Answer

Enthalpy change for a reaction (ΔH) is given by the expression,

ΔH = ΔU + ΔngRT

Where, ΔU = change in internal energy

Δng = change in number of moles

For the given reaction,

Δng = \sum  ng (products) - \sum  ng (reactants)

= (2 - 1.5) moles

Δng = 0.5 moles

And,

ΔU = -742.7 kJ mol-1

T = 298 K

R = 8.314 x 10-3 kJ mol-1 K-1

Substituting the values in the expression of ΔH:

ΔH = (-742.7 kJ mol-1) + (0.5 mol) (298 K) (8.314 x 10-3 kJ mol-1 K-1)

= -742.7 + 1.2

ΔH = -741.5 kJ mol-1

Popular Questions of Class Chemistry

Recently Viewed Questions of Class Chemistry

1 Comment(s) on this Question

Write a Comment: