For the following reaction at 50° C and at 2 atm pressure,...

Let the initial moles of $N_2O_5$ be $2$.

The given reaction is:
$2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$

Since $N_2O_5$ is $50\%$ dissociated (degree of dissociation $\alpha = 0.5$), the moles at equilibrium are:
Moles of $N_2O_5 = 2(1 - 0.5) = 1$
Moles of $N_2O_4 = 2(0.5) = 1$
Moles of $O_2 = 0.5$

Total moles at equilibrium $= 1 + 1 + 0.5 = 2.5$

Given the total pressure $P = 2$ atm, the partial pressures of the gases are:
$P_{N_2O_5} = \dfrac{1}{2.5} \times 2 = 0.8$ atm
$P_{N_2O_4} = \dfrac{1}{2.5} \times 2 = 0.8$ atm
$P_{O_2} = \dfrac{0.5}{2.5} \times 2 = 0.4$ atm

The equilibrium constant $K_p$ is:
$K_p = \dfrac{(P_{N_2O_4})^2 \cdot P_{O_2}}{(P_{N_2O_5})^2} = \dfrac{(0.8)^2 \times 0.4}{(0.8)^2} = 0.4$

The standard free energy change $\Delta G^\circ$ is given by:
$\Delta G^\circ = -RT \ln K_p$

Using the given values:
$\ln(0.4) = 2.303 \log(0.4) = 2.303 (\log 4 - \log 10)$
$\ln(0.4) = 2.303 (2 \log 2 - 1) = 2.303 (2 \times 0.30 - 1) = 2.303 (-0.4) = -0.9212$

Now, substituting the values into the $\Delta G^\circ$ equation ($T = 50^\circ C + 273 = 323$ K):
$\Delta G^\circ = -8.314 \times 323 \times (-0.9212)$
$\Delta G^\circ = 2473.81$ J mol$^{-1}$

The magnitude of the standard free energy change to the nearest integer is $2474$ J mol$^{-1}$.

Answer: $2474$