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consider the reaction cl₂(g) + br₂(g) → 2 brcl(g) at 25 °c. the entropy…

Question

consider the reaction cl₂(g) + br₂(g) → 2 brcl(g) at 25 °c. the entropy change for the reaction is 11.6 j/mol·k. if the standard enthalpy change is 29.4 kj/mol for the reaction, then what is the standard free energy for the reaction? | kj/mol

Explanation:

Step1: Recall the formula for Gibbs free energy change

The formula for the standard Gibbs free energy change ($\Delta G^\circ$) is $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$, where $\Delta H^\circ$ is the standard enthalpy change, $T$ is the temperature in Kelvin, and $\Delta S^\circ$ is the standard entropy change.

Step2: Convert temperature to Kelvin

The temperature is given as $25^\circ$C. To convert to Kelvin, we use $T = 25 + 273.15 = 298.15$ K.

Step3: Convert entropy change to kJ/mol·K

The entropy change ($\Delta S^\circ$) is given as $11.6$ J/mol·K. To convert to kJ/mol·K, we divide by 1000: $\Delta S^\circ = \frac{11.6}{1000} = 0.0116$ kJ/mol·K.

Step4: Substitute values into the formula

We know $\Delta H^\circ = 29.4$ kJ/mol, $T = 298.15$ K, and $\Delta S^\circ = 0.0116$ kJ/mol·K. Substituting into the formula:
$$\Delta G^\circ = 29.4 - (298.15 \times 0.0116)$$

Step5: Calculate the product $T\Delta S^\circ$

First, calculate $298.15 \times 0.0116$:
$298.15 \times 0.0116 \approx 3.45854$

Step6: Calculate $\Delta G^\circ$

Subtract the result from $\Delta H^\circ$:
$\Delta G^\circ = 29.4 - 3.45854 \approx 25.94146$

Answer:

25.9 (or approximately 25.94, depending on rounding)