For the reaction SbCl₅(g) $\leftrightharpoons$ SbCl₃(g) + Cl₂(g) ,
$\Delta$ G^$\circ$_f (SbCl₅) = -334.34 kJ/mol
$\Delta$ G^$\circ$_f (SbCl₃) = -301.25 kJ/mol
$\Delta$ H^$\circ$_f (SbCl₅) = -394.34 kJ/mol
$\Delta$ H^$\circ$_f (SbCl₃) = -313.80 kJ/mol
Calculate the value of the equilibrium constant (K_p) for this reaction at 298 K.

Question

For the reaction SbCl₅(g)

\leftrightharpoons

SbCl₃(g) + Cl₂(g) ,

\Delta

G^$\circ$_f (SbCl₅) = -334.34 kJ/mol

\Delta

G^$\circ$_f (SbCl₃) = -301.25 kJ/mol

\Delta

H^$\circ$_f (SbCl₅) = -394.34 kJ/mol

\Delta

H^$\circ$_f (SbCl₃) = -313.80 kJ/mol
Calculate the value of the equilibrium constant (K_p) for this reaction at 298 K.

Quizplus · Accepted Answer

The equilibrium constant (K_p) can be calculated using the formula: K_p = e (-ΔG°/RT), where R is the gas constant (8.314 J/mol·K), T is the temperature in kelvins, and ΔG° is the standard free energy change for the reaction. Converting the given values of ΔG° to joules and substituting in the formula, we get: K_p = e [(334340 - 301250) J/mol / (8.314 J/mol·K·298 K)] K_p = e [-16.278] K_p = 1.58 × 10^-7 Taking the antilogarithm of the result, we get: K_p = 1.58 × 10^1 Therefore, the best choice is B, as it is the only option that matches this value.

For the Reaction SbCl5(g) ⇋\leftrightharpoons⇋ SbCl3(g) + Cl2(g) Δ\DeltaΔ G ∘\circ∘ f (SbCl5) = -334

For the Reaction SbCl₅(g) $\leftrightharpoons$ SbCl₃(g) + Cl₂(g) $\Delta$ G^$\circ$_f (SbCl₅) = -334