Question

Calculate the entropy change accompanying the following change of state.

$H_{2}O(s,-{10}^{\circ }C,1atm)\to H_{2}O(l,{10}^{\circ }C,1atm)$
$C_P$ for ice = $9calde{g}^{-1}mo{l}^{-1}$

$C_P$ for $H_{2}O$ = $18calde{g}^{-1}mo{l}^{-1}$

Latent heat of fusion of ice = $1440calmo{l}^{-1}$ at $0^{\circ }C$

• A. $-8.88calde{g}^{-1}mo{l}^{-1}$
• B. $-7.77calde{g}^{-1}mo{l}^{-1}$
• C. $-6.3calde{g}^{-1}mo{l}^{-1}$
• D. None of these

Answer 1

The correct option is C

$-6.3calde{g}^{-1}mo{l}^{-1}$

The total process involves three stages and entropy change can be calculated for each stage as follows:

$[\Delta S_1\Rightarrow \text{For changing 1 mole of ice from -}{10}^{\circ }C,\text{1 atm to}{0}^{\circ }C‘1atm]$

$\Delta S_1={\int }_{-10}^{0}n.\frac{C_p}{T}dT$

= $n\times C_P\times 2.303\times log\frac{273}{263}$

= $1\times 9\times 2.303\times 0.0162$

= $0.336calde{g}^{-1}mo{l}^{-1}$

$\Delta S_2\Rightarrow$ for melting 1 mol of ice at $0^{\circ }C$ will be as

$\Delta S_2=\frac{q_{rev}}{T}$ = $\frac{1440}{273}$ = $5.25calde{g}^{-1}mo{l}^{-1}$

$\Delta S_3\Rightarrow$ for heating 1 mol of $H_{2}O$ from $0^{\circ }C$ to ${10}^{\circ }C$ at 1 atm

$\Delta S_3={\int }_0^{10}\frac{n{C}_p}{T}dT$

=$1\times 18\times 2.303log\frac{283}{273}$

= $0.647calde{g}^{-1}mo{l}^{-1}$

$\Delta S=0.336+5.276+0.647$

=$6.258calde{g}^{-1}mo{l}^{-1}$