Question

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

Answer 1

`K_(eq) '~~ 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 (`"T" = 298 color(white)(l) "K"`.)

Apply the van't Hoff equation:

`ln((K_2)/(K_1)) = (- Delta H^"o")/(R) * (1/(T_2) - 1/(T_1))`
#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 * 2.1 = 1.26`

Given that `K_1 = 2.1` at `"T" = 298 color(white)(l) "K"`.