MENU
Question -

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



Answer -

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

ΔH = ΔU + ΔngRT

Where,

ΔU = change in internalenergy

Δng = change in numberof 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 × 10–3 kJ mol–1 K–1

Substituting the values in the expression ofΔH:

ΔH = (–742.7 kJ mol–1) + (0.5 mol) (298 K)(8.314 × 10–3 kJ mol–1 K–1)

= –742.7 + 1.2

ΔH = –741.5 kJ mol–1

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×