Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of 
at equilibrium = 0.067 mol
Concentration of 
at equilibrium = 0.021 mol
Concentration of 
at equilibrium = 0.040 mol
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 424.3
Answer:
The rate of the reaction increased by a factor of 1012.32
Explanation:
Applying Arrhenius equation
ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)
where;
k₂/k₁ is the ratio of the rates which is the factor
Ea is the activation energy = 274 kJ/mol.
T₁ is the initial temperature = 231⁰C = 504 k
T₂ is the final temperature = 293⁰C = 566 k
R is gas constant = 8.314 J/Kmol
Substituting this values into the equation above;
ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)
ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)
ln(k₂/k₁) = 6.92
k₂/k₁ = exp(6.92)
k₂/k₁ = 1012.32
The rate of the reaction increased by 1012.32
a resource that cannot be replenished in a short period of time
The answer should be D all of the above
