Ncert Exercises & Solutions

Q.10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –

10.0°C. $∆_{f u s} H$ = 6.03 kJ $m o l^{- 1}$ at 0°C.
$C_{p} [H_{2} O (l)]$ = 75.3 J $m o l^{- 1}$ $K^{- 1}$
$C_{p} [H_{2} O (s)]$ = 36.8 J $m o l^{- 1}$ $K^{- 1}$

NEETprep Answer:

Total enthalpy change involved in the transformation is the sum of the following changes:
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

T o t a l ∆ H = C_{p} [H_{2} O C l] ∆ T + ∆ H_{f r e e z i n g} + C_{p} [H_{2} O_{(s)}] ∆ T

= (75.3 J $m o l^{- 1}$ $K^{- 1}$ ) (0 – 10)K + (–6.03 × 103 J $m o l^{- 1}$ ) + (36.8 J $m o l^{- 1}$ $K^{- 1}$ ) (–10 – 0)K

= –753 J $m o l^{- 1}$ – 6030 J $m o l^{- 1}$ – 368 J $m o l^{- 1}$
= –7151 J $m o l^{- 1}$
= –7.151 kJ $m o l^{- 1}$
Hence, the enthalpy change involved in the transformation is –7.151 kJ $m o l^{- 1}$ .

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