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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)


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)


Grade:11

3 Answers

Aman Bansal
592 Points
12 years ago

Dear Sha,

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 × 10–3 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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Rishi Sharma
askIITians Faculty 646 Points
3 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

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