Question

Question 6.8 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–1at 298 K. Calculate enthalpy change for the reaction at 298 K.

Class XI Thermodynamics Page 182

Answer 1

Concept :- we know, from first law of thermodynamics,
enthalpy change{∆H} = ∆U{internal energy} + ∆ngRT
Where,
∆ng = no of mole of gaseous products - no of mole of gaseous reactant
e.g., ∆ng = np - nr


reaction follows :
NH2CN(s) +(3/2)O2(g)-->N2(g)+CO2(g)

∆ng = np - nr
= (1 + 1) - 3/2 = 1/2 mol
= 0.5 mol

Now, ∆H = ∆U + ∆ngRT
= -742.7 KJ +(0.5mol × 8.314 J/mol/K × 298K)
= -742 KJ + 1238.786 J
= -742 KJ + 1.238786 KJ
= -741.46 KJ

Hence, enthalpy change for the reaction is 741.46 KJ for combustion of 1 mole of NH2CN.

Answer 2

Answer:-741.75kJmol^-1

Explanation:pls refer the attached pic