Question

consider the following reaction at 25 °c: 5 so₃(g) + 2 nh₃(g) → 2 no(g) + 5 so₂(g) + 3 h₂o(g) if δh° = 42.4 kj/mol and δs° = 562.3 j/mol·k, determine the standard free energy for the reaction at 25 °c.

Explanation:

Step1: Recall the formula for standard free energy change

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

Step2: Convert temperature to Kelvin

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

Step3: Convert $\Delta S^\circ$ units to kJ/mol·K

Given $\Delta S^\circ = 562.3\ J/mol\cdot K$. Since $1\ kJ = 1000\ J$, we divide by 1000: $\Delta S^\circ=\frac{562.3}{1000}=0.5623\ kJ/mol\cdot K$.

Step4: Substitute values into the formula

We know $\Delta H^\circ = 42.4\ kJ/mol$, $T = 298.15\ K$, and $\Delta S^\circ = 0.5623\ kJ/mol\cdot K$.
Substitute into $\Delta G^\circ=\Delta H^\circ - T\Delta S^\circ$:
$\Delta G^\circ=42.4\ kJ/mol-(298.15\ K\times0.5623\ kJ/mol\cdot K)$
First, calculate $298.15\times0.5623\approx298.15\times0.5623 = 167.6\ kJ/mol$ (approximate calculation).
Then, $\Delta G^\circ=42.4 - 167.6=- 125.2\ kJ/mol$ (more accurately, $298.15\times0.5623 = 298.15\times0.5 + 298.15\times0.06 + 298.15\times0.0023=149.075+17.889+0.685745 = 167.649745$)
So $\Delta G^\circ=42.4 - 167.649745=- 125.249745\ kJ/mol\approx - 125.2\ kJ/mol$ (or more precisely, - 125.2 kJ/mol when rounded appropriately).

Answer:

The standard free energy change $\Delta G^\circ$ is approximately $\boldsymbol{-125.2\ kJ/mol}$ (or more precisely, around - 125 kJ/mol depending on rounding during calculation).