Question:
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) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l)

Answer:

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

ΔH = ΔU + Δn_gRT

Where, ΔU = change in internal energy

Δng = change in number of moles

For the given reaction,

Δn_g = $\sum$ ng (products) - $\sum$ n_g (reactants)

= (2 - 1.5) moles

Δn_g = 0.5 moles

And,

ΔU = -742.7 kJ mol^-1

T = 298 K

R = 8.314 x 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 x 10^-3 kJ mol^-1 K^-1)

= -742.7 + 1.2

ΔH = -741.5 kJ mol^-1

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Question:
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) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l)

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Question: 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) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(l)

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Question:
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) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l)