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)
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 = ng (products) -
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
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and &Delt....
Comments
Thanks a lot