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  • Calculate the entropy change  when 2.00 moles of ice at <img alt="\math

Calculate the entropy change \mathrm{(\Delta S)} when 2.00 moles of ice at \mathrm{-10^{\circ} \mathrm{C}} is heated to steam at \mathrm{150^{\circ} \mathrm{C}.} The heat capacities

\mathrm{\left(C_p\right)} for ice, liquid water, and steam are \mathrm{37.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, 75.3 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K},} and \mathrm{33.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}, respectively.

Option: 1

\mathrm{164.5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}


Option: 2

\mathrm{164.5 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}


Option: 3

\mathrm{175 \mathrm{~J} / \mathrm{mol}-\mathrm{K}}


Option: 4

\mathrm{\text { -113. J/mol-K }}


Answers (1)

best_answer

Given data:

\mathrm{ \begin{aligned} & n=2.00 \mathrm{~mol} \\\\ & C_{p, \text { ice }}=37.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, \quad C_{p, \text { water }}=75.3 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}, \quad C_{p, \text { steam }}=33.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\\\ & \Delta H_{\text {fusion }}=6.01 \times 10^3 \mathrm{~J} / \mathrm{mol}, \quad \Delta H_{\text {vaporization }}=40.79 \times 10^3 \mathrm{~J} / \mathrm{mol} \\\\ & T_1=-10^{\circ} \mathrm{C}=263.15 \mathrm{~K}, \quad T_2=150^{\circ} \mathrm{C}=423.15 \mathrm{~K} \end{aligned} }

We'll calculate the entropy change for each phase transition and temperature range:

a. Entropy change for heating ice from \mathrm{-10^{\circ} \mathrm{C}\, \, to \, \, 0^{\circ} \mathrm{C} :}

\mathrm{ \Delta S_{\text {ice }}=\int_{263.15 \mathrm{~K}}^{273.15 \mathrm{~K}} \frac{C_{p, \text { ice }}}{T} d T }

Using the integral formula \mathrm{\int \frac{C_p}{T} d T=C_p \ln \left(\frac{T_2}{T_1}\right) :}

\mathrm{ \Delta S_{\text {ice }}=C_{p, \text { ice }} \ln \left(\frac{273.15 \mathrm{~K}}{263.15 \mathrm{~K}}\right)=37.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 0.0387=1.38 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} }

b. Entropy change for melting ice at \mathrm{0^{\circ} \mathrm{C} :}

\mathrm{ \Delta S_{\text {fusion }}=\frac{\Delta H_{\text {fusion }}}{T}=\frac{6.01 \times 10^3 \mathrm{~J} / \mathrm{mol}}{273.15 \mathrm{~K}}=22.02 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} }

c. Entropy change for heating liquid water from \mathrm{0^{\circ} \mathrm{C}} to \mathrm{100^{\circ} \mathrm{C} :}

                       \mathrm{ \Delta S_{\text {water }}=\int_{273.15 \mathrm{~K}}^{373.15 \mathrm{~K}} \frac{C_{p, \text { water }}}{T} d T }

Using the integral formula:
\mathrm{ \Delta S_{\text {water }}=C_{p, \text { water }} \ln \left(\frac{373.15 \mathrm{~K}}{273.15 \mathrm{~K}}\right)=75.3 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 0.3665=27.59 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} }
d. Entropy change for vaporizing water at \mathrm{100^{\circ} \mathrm{C} :}

\mathrm{ \Delta S_{\text {vaporization }}=\frac{\Delta H_{\text {vaporization }}}{T}=\frac{40.79 \times 10^3 \mathrm{~J} / \mathrm{mol}}{373.15 \mathrm{~K}}=109.20 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} }

e. Entropy change for heating steam from \mathrm{100^{\circ} \mathrm{C}}  to \mathrm{150^{\circ} \mathrm{C} :}

                                \mathrm{ \Delta S_{\text {steam }}=\int_{373.15 \mathrm{~K}}^{423.15 \mathrm{~K}} \frac{C_{p, s t e a m}}{T} d T }
Using the integral formula:
\mathrm{ \Delta S_{\text {steam }}=C_{p, \text { steam }} \ln \left(\frac{423.15 \mathrm{~K}}{373.15 \mathrm{~K}}\right)=33.6 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times 0.1338=4.49 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} }
Now sum up the entropy changes for each phase transition and temperature range:

\mathrm{ \begin{gathered} \Delta S=\Delta S_{\text {ice }}+\Delta S_{\text {fusion }}+\Delta S_{\text {water }}+\Delta S_{\text {vaporization }}+\Delta S_{\text {stcam }} \\\\ \Delta S=1.45 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+22.02 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+27.59 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+109.20 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}+4.49 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}=164.75 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{gathered} }
The correct option is B).

Posted by

sudhir kumar

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