Grade 11Physical Chemistry

Q6.8 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\2O2(g) → N2(g) + CO2(g) + H2O(l)

sha swa

14 Years agoGrade 11

3 Answers

Aman Bansal

14 Years ago

Dear Sha,

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

ΔH = ΔU + Δn_gRT

Where,

ΔU = change in internal energy

Δn_g = change in number of moles

For the given reaction,

Δn_g = ∑n_g (products) – ∑n_g (reactants)

= (2 – 2.5) moles

Δn_g = –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 of ΔH:

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 = –743.9 kJ mol^–1

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Aman Bansal

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6 Years ago

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Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

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 – 2.5) moles
Δng= –0.5 moles
And,
ΔU= –742.7 kJ mol–1
T= 298 K
R = 8.314 × 10^–3kJ mol–1K–1
Substituting the values in the expression of ΔH:
ΔH= (–742.7 kJ mol–1) + (–0.5 mol) (298 K) (8.314 × 10^–3kJ mol–1K–1)
= –742.7 – 1.2
ΔH= –743.9 kJ mol–1

Thanks and Regards