The reaction of cyanamide, NH 2 CN (s), with dioxygen was carried out in a bomb calorimeter, and D U was found to be –742.7 kJ mol –1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

Question

Philoid · Accepted Answer

We know that,

ΔU = ΔH - Δn_gRT

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,

Δn_g = 2 – 3/2 = 1/2

∴ ΔU = ΔH - Δn_gRT

Given,

ΔH = -742.7 kJ

R =8.314 J K^-1 mol^-1

= 8.414 × 10^-3 kJ K^-1 mol^-1

Δn_g =1/2

T = 298 K

Thus,

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

= -742.7 + 1.239

= -741.46 kJ

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.

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.