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The Reaction of Cyanamide, Nh2cn(S),With Dioxygen Was Carried Out in a Bomb Calorimeter, and δUwas Found to Be –742.7 Kj Mol–1at 298 K. Calculate Enthalpy Change for the Reaction at 298 K. - CBSE (Science) Class 11 - Chemistry

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Question

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

`NH_2 CN(g) + 3/2 O_2(g) -> N_2(g) + CO_2(g) + H_2O(1)`

Solution 1

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 × 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 = –741.5 kJ mol–1

Solution 2

∆U = – 742.7 KJ-1  mol-1 ; ∆ng = 2 – 3/2 = + 1/2 mol.

R = 8.314 x 10-3KJ-1  mol-1 ; T = 298 K

According to the relation,∆H = ∆U+∆ngRT

∆H = (- 742.7 kj) + (1/2 mol) x (8.314 x10-3 KJ-1  mol-1 ) x (298 K)

= – 742.7 kj + 1.239 kj = – 741.5 kj.

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Solution The Reaction of Cyanamide, Nh2cn(S),With Dioxygen Was Carried Out in a Bomb Calorimeter, and δUwas Found to Be –742.7 Kj Mol–1at 298 K. Calculate Enthalpy Change for the Reaction at 298 K. Concept: Enthalpy Change, ∆_rH of a Reaction - Reaction Enthalpy - Standard Enthalpy of Reactions.
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