Consider the reaction

at 25°C, for which the following data are relevant:
-Calculate $\Delta$G° for the reaction at 25°C.

Question

Consider the reaction
    
at 25°C, for which the following data are relevant: 
-Calculate $\Delta$G° for the reaction at 25°C.

Quizplus · Accepted Answer

To calculate the standard Gibbs free energy change of the reaction ($\Delta G^\circ$), we use the equation:
$$\Delta G^\circ = -RT\ln K$$
where $T$ is the temperature in Kelvin, $R$ is the gas constant, and $K$ is the equilibrium constant of the reaction.

First, we need to balance the equation:
$$2NS_{2}O_{5} (g) + 4O_{2} (g) ightarrow 4NO_{2} (g) + 2S_{2}O_{6} (g)$$

Now we can write the expression for the equilibrium constant:
$$K = \frac{(P_{NO_{2}})^4 (P_{S_{2}O_{6}})^2}{(P_{NS_{2}O_{5}})^2 (P_{O_{2}})^4}$$
where $P$ represents the partial pressure of each species.

From the given data, we know the following equilibrium pressures:
$P_{NS_{2}O_{5}} = 0.100$ atm, $P_{NO_{2}} = 0.520$ atm, $P_{S_{2}O_{6}} = 0.268$ atm, and $P_{O_{2}} = 0.640$ atm.

Substituting these values into the equation for $K$, we get:
$$K = \frac{(0.520)^4 (0.268)^2}{(0.100)^2 (0.640)^4} = 2.72 	imes 10^6$$

Now we can calculate the standard Gibbs free energy change:
$$\Delta G^\circ = -RT\ln K = -(8.314 	ext{ J/mol}\cdot	ext{K})(298 	ext{ K})\ln(2.72	imes 10^6) = -25.2 	ext{ kJ/mol}$$

Consider the Reaction at 25°C Δ\DeltaΔ

Consider the Reaction

at 25°C $\Delta$