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valkas [14]
3 years ago
5

Consider the reaction to produce methanolCO(g) + 2H2 (g) <-----> CH3OHAn equilibrium mixture in a 2.00-L vessel is found t

o contain 0.0406 mol CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K.Calculate Kc at this temperature.Does equilibrium favor reactants or products?
Chemistry
1 answer:
MariettaO [177]3 years ago
7 0

Answer : The value of K_c of the reaction is 10.5 and the reaction is product favored.

Explanation : Given,

Moles of CH_3OH at equilibrium = 0.0406 mole

Moles of CO at equilibrium = 0.170 mole

Moles of H_2 at equilibrium = 0.302 mole

Volume of solution = 2.00 L

First we have to calculate the concentration of CH_3OH,CO\text{ and }H_2 at equilibrium.

\text{Concentration of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Volume of solution}}=\frac{0.0406mole}{2.00L}=0.0203M

\text{Concentration of }CO=\frac{\text{Moles of }CO}{\text{Volume of solution}}=\frac{0.170mole}{2.00L}=0.085M

\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.302mole}{2.00L}=0.151M

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)

The expression of equilibrium constant K_c for the reaction will be:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the values in this expression, we get :

K_c=\frac{(0.0203)}{(0.085)\times (0.151)^2}

K_c=10.5

Therefore, the value of K_c of the reaction is, 10.5

There are 3 conditions:

When K_{c}>1; the reaction is product favored.

When K_{c}; the reaction is reactant favored.

When K_{c}=1; the reaction is in equilibrium.

As the value of K_{c}>1. So, the reaction is product favored.

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