Ncert Exercises & Solutions

Q.15. Calculate the enthalpy change for the process

$C C l_{4 (g)} \to C_{(g)} + 4 C l_{(g)}$ and calculate bond enthalpy of C-Cl in $C C l_{4 (g)}$ .

$∆_{v a p} H^{θ} (C C l_{4}) = 30.5 k J m o l^{- 1} .$
$∆_{f} H^{θ} (C C l_{4}) = 135.5 k J m o l^{- 1} .$
$∆_{a} H^{θ} (C) = 715.0 k J m o l^{- 1}, w h e r e ∆_{a} H^{θ} i s e n t h a l p y o f a t o m i s a t i o n$
$∆_{a} H^{θ} (C l_{2}) = 242 k J m o l^{- 1}$

NEETprep Answer:

The chemical equations implying to the given values of enthalpies are:

(i) C C l_{4 (l)} \to C C {l_{4 (g)}}_{∆ v a p} H^{θ} = 30.5 k J m o l^{- 1} (i i) C_{(s)} \to C_{(g)} a^{θ} ∆ H = 715.0 k J m o l^{- 1} (i i i) C l_{2 (g)} \to 2 C l_{(g)} a^{θ} ∆ H = 242 k J m o l^{- 1} (i v) C_{(g)} + 4 C l_{(g)} \to C C l_{4 (g)} ∆ r H = - 135.5 k J m o l^{- 1}

Enthalpy change for the given process

C C l_{4 (g)} \to C_{(g)} + 4 C l_{(g)}

can be calculated using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

∆ H = ∆ a H^{θ} (C) + 2 ∆ a H^{θ} (C l_{2}) - ∆ v a p H^{θ} - ∆ f H = (715.0 k J m o l^{- 1}) + 2 (242 k J m o l^{- 1}) - (30.5 k J m o l^{- 1}) - (- 135.5 k J m o l^{- 1}) ∴ ∆ H = 1304 k J m o l^{- 1}

Bond enthalpy of C-Cl bond in

C C l_{4 (g)}

\frac{1304}{4} k J m o l^{- 1} = 326 k J m o l^{- 1}