For the reaction PCl₅(g)→ PCl₃(g)+ Cl₂(g)at 298 K,K_eq = 1.87 × 10^-7,Δ_rS° = 181.92 J/mol K,what is K_eq at 200 K if Δ_rH° = 92.6 kJ/mol? A)2.08 × 10^-15 B)16.8 C)1.87 × 10^-7 D)5.35 × 10⁶ E)5.95 × 10^-2

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To find the new equilibrium constant ($K_{eq}$) at 200 K, we can use the van't Hoff equation which relates the change in the equilibrium constant with temperature:$$\ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$Given:- $K_1 = 1.87 \times 10^{-7}$ at $T_1 = 298 K$- $\Delta H^\circ = 92.6 kJ/mol = 92600 J/mol$- $T_2 = 200 K$We need to find $K_2$.Rearranging the van't Hoff equation to solve for $K_2$:$$K_2 = K_1 \exp\left(-\frac{\Delta H^\circ}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\right)$$Substituting the values (with $R = 8.314 J/(mol\cdot K)$):$$K_2 = 1.87 \times 10^{-7} \exp\left(-\frac{92600}{8.314}\left(\frac{1}{200} - \frac{1}{298}\right)\right)$$After calculating, you'll find that $K_2$ significantly decreases, which matches the option:A) $2.08 \times 10^{-15}$

For the Reaction PCl5(g)→ PCl3(g)+ Cl2(g)at 298 K,Keq = 1

For the Reaction PCl₅(g)→ PCl₃(g)+ Cl₂(g)at 298 K,K_eq = 1