The equilibrium constant, kc, for the following reaction is 9.52×10-2 at 350 k. ch4 (g) + ccl4 (g) 2 ch2cl2 (g) calculate the equilibrium concentrations of reactants and product when 0.377 moles of ch4 and 0.377 moles of ccl4 are introduced into a 1.00 l vessel at 350 k.

The equilibrium concentration of is 0.377 M and equilibrium concentration of is 0.116 M

Explanation:

To calculate the molarity of solution, we use the equation:

Volume of solution = 1 L

Moles of methane = 0.377 moles

Putting values in above equation, we get:

Volume of solution = 1 L

Moles of carbon tetrachloride = 0.377 moles

Putting values in above equation, we get:

For the given chemical equation:

The expression for for the given equation follows:

We are given:

Putting values in above equation, we get:

Hence, the equilibrium concentration of is 0.377 M and equilibrium concentration of is 0.116 M

i don t now

the answer is d.) potassium