Answer to Question #297160 in Chemistry for Victoria

Question

I) In an equilibrium mixture at 100°C the following gas concentrations were found:

CO : 3.76 x 10-3 mol L-1 H2: 4.30 x 10-3 mol L-1

CH3OH: 4.17 x 10-10 mol L-1

Calculate the value of K(eq) at 100°C

Answer 1

Solution:

CO(g) + 2H₂(g) ⇌ CH₃OH(g) + 52 kJ

The equilibrium constant (K_eq) expression is as follows:

[CH₃OH] = 4.17×10^-10 M

[CO] = 3.76×10^-3 M

[H₂] = 4.30×10^-3 M

Therefore,

Thus,

K_eq = 6.00×10^-3 M^-2 = 6.00×10^-3 L² mol^-2

Answer: The value of K_eq is 6.00×10^-3 L² mol^-2