A chemical reaction has \ \ DeltaH^{circ}=-5.8\ text{kJmol}^{-1]\ \  and an equilibrium constant of \ \ 2.1\ \  at \ \ T=298\ K.\ \ ...?

A chemical reaction has $\setminus \setminus \Delta {H}^{\circ} = - 5.8 \setminus {\textrm{k J m o l}}^{- 1} \setminus \setminus$ and an equilibrium constant of $\setminus \setminus 2.1 \setminus \setminus$ at $\setminus \setminus T = 298 \setminus K . \setminus \setminus$...?

Jul 18, 2018

${K}_{e q} ' \approx 1.26$

Explanation:

As @truong-son-n has mentioned, the van't Hoff equation relates the equilibrium constant to temperature given the enthalpy change of the reaction under standard conditions ($\text{T" = 298 color(white)(l) "K}$.)

Apply the van't Hoff equation:

$\ln \left(\frac{{K}_{2}}{{K}_{1}}\right) = \frac{- \Delta {H}^{\text{o}}}{R} \cdot \left(\frac{1}{{T}_{2}} - \frac{1}{{T}_{1}}\right)$
color(white)(ln((K_2)/(K_1))) = -(-5.8 * 10^3 color(white)(l) color(navy)("J") * mol^(-1)) /(8.314 color(white)(l) color(navy)("J") * mol^(-1) * "K"^(-1)) * (1/(382.0 color(white)(l) "K") - 1/(298.0 color(white)(l) "K"))
color(white)(ln((K_2)/(K_1))) = ‐0.515

(K_2) / (K_1) = e^(‐0.515) ~~0.598

Therefore

${K}_{2} = 0.598 \cdot 2.1 = 1.26$

Given that ${K}_{1} = 2.1$ at $\text{T" = 298 color(white)(l) "K}$.