Magnitude of Standard Entropy Change for Chemical Reaction

To find the magnitude of the standard entropy change ($\Delta S^\theta$) for the reaction, we use the standard thermodynamic relations:

$\Delta G^\theta = -RT \ln K$ $\Delta G^\theta = \Delta H^\theta - T \Delta S^\theta$

Equating the two expressions for standard Gibbs free energy change ($\Delta G^\theta$):

$\Delta H^\theta - T \Delta S^\theta = -RT \ln K$

Rearranging the equation to solve for $\Delta S^\theta$:

$\Delta S^\theta = \frac{\Delta H^\theta}{T} + R \ln K$

Step 1: Calculate $\ln K$

We know that $\ln K = \ln(10) \times \log_{10}(K)$. First, let me find $\log_{10}(K)$:

$K = 1.8 \times 10^{-7} = \frac{18}{10} \times 10^{-7}$

$\log_{10}(K) = \log_{10}(18) - \log_{10}(10) - 7$ $\log_{10}(18) = \log_{10}(3^2 \times 2) = 2\log_{10}(3) + \log_{10}(2)$

Given values:

• $\log_{10}(3) = 0.48$
• $\log_{10}(2) = 0.30$
• $\ln(10) = 2.3$

Substituting these into the expression:

$\log_{10}(18) = 2(0.48) + 0.30 = 0.96 + 0.30 = 1.26$ $\log_{10}(K) = 1.26 - 1 - 7 = -6.74$

Now, calculating $\ln K$:

$\ln K = 2.3 \times (-6.74) = -15.502$

Step 2: Calculate $\Delta S^\theta$

Given:

• $\Delta H^\theta = 28.40 \text{ kJ mol}^{-1} = 28400 \text{ J mol}^{-1}$
• $T = 300 \text{ K}$
• $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Substitute the known values into the equation for $\Delta S^\theta$:

$\Delta S^\theta = \frac{28400}{300} + 8.3 \times (-15.502)$

$\frac{28400}{300} = \frac{284}{3} \approx 94.6667 \text{ J K}^{-1} \text{ mol}^{-1}$ $R \ln K = 8.3 \times (-15.502) = -128.6666 \text{ J K}^{-1} \text{ mol}^{-1}$

$\Delta S^\theta = 94.6667 - 128.6666 = -33.9999 \text{ J K}^{-1} \text{ mol}^{-1}$

Step 3: Find the magnitude

$|\Delta S^\theta| = 33.9999 \text{ J K}^{-1} \text{ mol}^{-1} \approx 34 \text{ J K}^{-1} \text{ mol}^{-1}$