Ncert Exercises & Solutions

Q.8. The reaction of cyanamide, $N H_{2} C N_{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ $m o l^{- 1}$ at 298 K. Calculate enthalpy change for the reaction at 298 K.

$N H_{2} C N_{(g)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + C O_{2 (g)} + H_{2} O_{(l)}$

NEETprep Answer:

Enthalpy change for a reaction (∆H) is given by the expression,
∆H = ∆U + ∆

n_{g} R T

Where,
∆U = change in internal energy
∆

n_{g}

= change in number of moles
For the given reaction,
∆ng = ∑

n_{g}

(products) – ∑

n_{g}

(reactants)
= (2 – 2.5) moles
∆ng = –0.5 moles
And,
∆U = –742.7 kJ

m o l^{- 1}

T = 298 K
R = 8.314 × 10–3 kJ

m o l^{- 1}

Substituting the values in the expression of ∆H:
∆H = (–742.7 kJ $m o l^{- 1}$ ) + (–0.5 mol) (298 K) (8.314 × $10^{- 3}$ kJ $m o l^{- 1}$ $K^{- 1}$ ) = –742.7 – 1.2
∆H = –743.9 kJ $m o l^{- 1}$

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