Zinc melts at $$ 420^{\circ} \mathrm{C} $$ and its standard entropy at $$ 25^{\circ} \mathrm{C} $$ is $$ 9.95 \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$. The standard entropy of zinc at $$ 750^{\circ} \mathrm{C} $$ is __ $$ \mathrm{Cal} / \mathrm{mol} / \mathrm{K} $$, round off to one decimal place). Given: Heat of fusion of $$ \mathrm{Zn}\left(\Delta \mathrm{H}_{1} ight)=1.74 \mathrm{kcal} / \mathrm{mol} $$ $$ C_{p}=5.35+2.40 imes 10^{-3} \mathrm{~T} \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$ $$ C_{p} $$ (liquid Zn$$ )=7.50 \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$

Question

Zinc melts at $$ 420^{\circ} \mathrm{C} $$ and its standard entropy at $$ 25^{\circ} \mathrm{C} $$ is $$ 9.95 \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$. The standard entropy of zinc at $$ 750^{\circ} \mathrm{C} $$ is __ $$ \mathrm{Cal} / \mathrm{mol} / \mathrm{K} $$, round off to one decimal place). Given: Heat of fusion of $$ \mathrm{Zn}\left(\Delta \mathrm{H}_{1}ight)=1.74 \mathrm{kcal} / \mathrm{mol} $$ $$ C_{p}< $$ solid $$ \mathrm{Zn}>=5.35+2.40 	imes 10^{-3} \mathrm{~T} \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$ $$ C_{p} $$ (liquid Zn$$ )=7.50 \mathrm{cal} / \mathrm{mol} / \mathrm{K} $$

UpStudy AI · Accepted Answer

To determine the standard entropy of zinc at $$ 750^{\circ} \mathrm{C} $$ (which is $$ 1023 \, \mathrm{K} $$), we'll follow these steps:

1. **Identify Phases and Temperature Ranges:**
   - **Solid Phase:** From $$ 298 \, \mathrm{K} $$ (25°C) to the melting point at $$ 693 \, \mathrm{K} $$ (420°C).
   - **Liquid Phase:** From $$ 693 \, \mathrm{K} $$ to $$ 1023 \, \mathrm{K} $$ (750°C).

2. **Calculate Entropy Changes:**

**a. Solid Phase ($$ 298 \, \mathrm{K} $$ to $$ 693 \, \mathrm{K} $$):**
   
   The heat capacity for solid zinc is given by:
   $$
   C_p^\text{solid} = 5.35 + 2.40 \times 10^{-3} T \, \mathrm{cal/mol/K}
   $$
   
   The entropy change ($$ \Delta S_\text{solid} $$) is:
   $$
   \Delta S_\text{solid} = \int_{298}^{693} \frac{C_p^\text{solid}}{T} \, dT = 5.35 \ln\left(\frac{693}{298}\right) + 0.0024 \times (693 - 298) \approx 4.502 + 0.948 = 5.450 \, \mathrm{cal/mol/K}
   $$
   
   **b. Phase Change (Melting at $$ 693 \, \mathrm{K} $$):**
   
   The entropy of fusion is:
   $$
   \Delta S_\text{fusion} = \frac{\Delta H_\text{fusion}}{T_m} = \frac{1740 \, \mathrm{cal/mol}}{693 \, \mathrm{K}} \approx 2.510 \, \mathrm{cal/mol/K}
   $$
   
   **c. Liquid Phase ($$ 693 \, \mathrm{K} $$ to $$ 1023 \, \mathrm{K} $$):**
   
   The heat capacity for liquid zinc is:
   $$
   C_p^\text{liquid} = 7.50 \, \mathrm{cal/mol/K}
   $$
   
   The entropy change ($$ \Delta S_\text{liquid} $$) is:
   $$
   \Delta S_\text{liquid} = \int_{693}^{1023} \frac{C_p^\text{liquid}}{T} \, dT = 7.50 \ln\left(\frac{1023}{693}\right) \approx 7.50 \times 0.3923 = 2.942 \, \mathrm{cal/mol/K}
   $$

3. **Sum All Contributions:**
   
   Starting with the standard entropy at $$ 298 \, \mathrm{K} $$:
   $$
   S(298 \, \mathrm{K}) = 9.95 \, \mathrm{cal/mol/K}
   $$
   
   Adding all entropy changes:
   $$
   S(1023 \, \mathrm{K}) = 9.95 + 5.450 + 2.510 + 2.942 = 20.852 \, \mathrm{cal/mol/K}
   $$
   
   Rounding to one decimal place:
   $$
   S(750^{\circ} \mathrm{C}) \approx 20.9 \, \mathrm{cal/mol/K}
   $$

**Final Answer:**

At $$ 750^{\circ} \mathrm{C} $$, the standard entropy of zinc is **20.9 cal mol⁻¹ K⁻¹**.
At $$ 750^{\circ} \mathrm{C} $$, the standard entropy of zinc is **20.9 cal/mol/K**.

Zinc melts at and its standard entropy at is . The standard entropy of zinc at is __ , round
off to one decimal place).
Given:
Heat of fusion of
solid
(liquid Zn

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Chemistry Questions

Zinc melts at and its standard entropy at is . The standard entropy of zinc at is __ , round off to one decimal place). Given: Heat of fusion of solid (liquid Zn