Let's solve each part step by step. **Table B2** is commonly found in thermodynamics or chemical engineering textbooks and provides polynomial expressions for heat capacities (\( C_p \)) as functions of temperature. --- ## a) **\( C_p \) of Ethane as a Function of Temperature** **Step 1: Write the general polynomial form from Table B2** - Usually, the heat capacity is given as: \[ C_p = a + bT + cT^2 + dT^3 \] where: - \( C_p \) is in J/(mol·K) or cal/(mol·K) - \( T \) is temperature in Kelvin **Step 2: Find the coefficients for ethane (\( C_2H_6 \))** - From a typical Table B2 (for gases, 298–150 K): \[ C_p^\circ (\text{C}_2\text{H}_6, g) = 19.502 + 5.024 \times 10^{-2}T - 1.765 \times 10^{-5}T^2 + 2.225 \times 10^{-9}T^3 \] (This is a standard value; please check your specific table if it differs.) **Step 3: State the applicable temperature range** - Usually valid for **298 K to 150 K**. **Final Answer:** \[ C_p^\circ (\text{C}_2\text{H}_6, g) = 19.502 + 5.024 \times 10^{-2}T - 1.765 \times 10^{-5}T^2 + 2.225 \times 10^{-9}T^3 \] Where \( T \) is in K, and **valid for \( 298 \leq T \leq 150 \) K**. --- ## b) **Enthalpy Change for 5 moles of Ethane (200°C to 350°C)** **Step 1: Convert °C to K** - \( T_1 = 200^\circ\text{C} = 473 \) K - \( T_2 = 350^\circ\text{C} = 623 \) K **Step 2: Write the enthalpy change formula** \[ \Delta H = n \int_{T_1}^{T_2} C_p \, dT \] where \( n = 5 \) moles. **Step 3: Integrate the \( C_p \) expression** \[ \int_{T_1}^{T_2} (a + bT + cT^2 + dT^3) dT = a(T_2-T_1) + \frac{b}{2}(T_2^2 - T_1^2) + \frac{c}{3}(T_2^3 - T_1^3) + \frac{d}{4}(T_2^4 - T_1^4) \] **Step 4: Plug in the coefficients and limits** - \( a = 19.502 \) - \( b = 5.024 \times 10^{-2} \) - \( c = -1.765 \times 10^{-5} \) - \( d = 2.225 \times 10^{-9} \) \[ \Delta H = 5 \left[ 19.502(623-473) + \frac{5.024 \times 10^{-2}}{2}(623^2-473^2) + \frac{-1.765 \times 10^{-5}}{3}(623^3-473^3) + \frac{2.225 \times 10^{-9}}{4}(623^4-473^4) \right] \] **Step 5: Calculate each term** - \( (623-473) = 150 \) - \( (623^2-473^2) = (623+473)(623-473) = 1096 \times 150 = 164,400 \) - \( (623^3-473^3) = (623-473)[(623)^2 + 623 \times 473 + (473)^2] = 150 \times (388,129 + 294,079 + 223,729) = 150 \times 905,937 = 135,890,550 \) - \( (623^4-473^4) \) is large, so let's use a calculator: - \( 623^4 = 150,668,161 \) - \( 473^4 = 50,102,641 \) - Difference: \( 100,565,520 \) Now, plug values in: \[ \Delta H = 5 \left[ 19.502 \times 150 + \frac{5.024 \times 10^{-2}}{2} \times 164,400 + \frac{-1.765 \times 10^{-5}}{3} \times 135,890,550 + \frac{2.225 \times 10^{-9}}{4} \times 100,565,520 \right] \] Now, calculate each term: 1. \( 19.502 \times 150 = 2,925.3 \) 2. \( .05024/2 = .02512; \ \ .02512 \times 164,400 = 4,129.73 \) 3. \( -1.765 \times 10^{-5}/3 = -5.883 \times 10^{-6}; \ -5.883 \times 10^{-6} \times 135,890,550 = -799.56 \) 4. \( 2.225 \times 10^{-9}/4 = 5.5625 \times 10^{-10}; \ 5.5625 \times 10^{-10} \times 100,565,520 = .056 \) Sum inside brackets: \[ 2,925.3 + 4,129.73 - 799.56 + .056 = 6,255.53 \] Multiply by 5: \[ \Delta H = 5 \times 6,255.53 = \boxed{31,277.65 \ \text{J}} \] Or in kJ: \[ \Delta H = 31.3 \ \text{kJ} \] --- ## c) **Enthalpy Change for 5 moles of Magnesium (400°C to 450°C)** **Step 1: Use the correct \( C_p \) for solid magnesium (from Table B2, usually for solids)** - Typical form: \[ C_p = a + bT + cT^2 \] For solid Mg: \[ C_p (\text{Mg}, \text{s}) = 24.98 + 1.55 \times 10^{-2}T - 3.30 \times 10^{-6}T^2 \] (Check your table for exact values.) **Step 2: Convert °C to K** - \( T_1 = 400^\circ\text{C} = 673 \) K - \( T_2 = 450^\circ\text{C} = 723 \) K **Step 3: Integrate** \[ \int_{T_1}^{T_2} (a + bT + cT^2) dT = a(T_2-T_1) + \frac{b}{2}(T_2^2-T_1^2) + \frac{c}{3}(T_2^3-T_1^3) \] **Step 4: Plug in coefficients and limits** - \( a = 24.98 \) - \( b = 1.55 \times 10^{-2} \) - \( c = -3.30 \times 10^{-6} \) \[ \Delta H = 5 \left[ 24.98 \times (723-673) + \frac{1.55 \times 10^{-2}}{2} (723^2-673^2) + \frac{-3.30 \times 10^{-6}}{3} (723^3-673^3) \right] \] **Calculate each term:** - \( (723-673) = 50 \) - \( (723^2-673^2) = (723+673) \times 50 = 1,396 \times 50 = 69,800 \) - \( (723^3-673^3) = (723-673)[(723)^2 + 723 \times 673 + (673)^2] = 50 \times (522,729 + 486,579 + 453,929) = 50 \times 1,463,237 = 73,161,850 \) Plug in: 1. \( 24.98 \times 50 = 1,249 \) 2. \( .0155/2 = .00775; \ .00775 \times 69,800 = 540.95 \) 3. \( -3.30 \times 10^{-6}/3 = -1.1 \times 10^{-6}; \ -1.1 \times 10^{-6} \times 73,161,850 = -80.478 \) Sum: \[ 1,249 + 540.95 - 80.478 = 1,709.47 \] Multiply by 5: \[ \Delta H = 5 \times 1,709.47 = \boxed{8,547.35 \ \text{J}} \] Or in kJ: \[ \Delta H = 8.55 \ \text{kJ} \] --- # **Summary of Answers** ### a) **\( C_p \) of ethane:** \[ C_p^\circ (\text{C}_2\text{H}_6, g) = 19.502 + 5.024 \times 10^{-2}T - 1.765 \times 10^{-5}T^2 + 2.225 \times 10^{-9}T^3 \] - with \( T \) in K, valid from **298 K to 150 K**. ### b) **Enthalpy change for 5 moles ethane (200°C → 350°C):** \[ \Delta H = \boxed{31,277.65 \ \text{J}} \approx 31.3 \ \text{kJ} \] ### c) **Enthalpy change for 5 moles magnesium (400°C → 450°C):** \[ \Delta H = \boxed{8,547.35 \ \text{J}} \approx 8.55 \ \text{kJ} \] --- **Note:** Always use your course's specific Table B2 for coefficients. If you provide the table, I can verify the numbers for you!