Consider the following reaction and its equilibrium constant: 4 cuo(s) + ch4(g) ⇌ co2(g) + 4 cu(s) + 2 h2o(g) kc = 1.10.
a reaction mixture contains 0.22 m ch4, 0.70 m co2 and 1.5 m h2o. which of the following statements is true concerning this system? a. the reaction will shift in the direction of products. b. the equilibrium constant will increase. c. the reaction will shift in the direction of reactants. d. the system is at equilibrium.

The answer for the question related to the reaction 4CuO(s) + CH₄(g) ⇄ CO₂(g) + 4Cu(s) + 2H₂O(g) with an equilibrium constant equal to 1.10, is that the reaction will shift in the direction of reactants (option C) when the concentration of CH₄, CO₂, and H₂O is 0.22, 0.70, and 1.5 M, respectively.

The reaction is:

4CuO(s) + CH₄(g) ⇄ CO₂(g) + 4Cu(s) + 2H₂O(g) (1)  

The equilibrium constant of the above reaction is Kc = 1.10.

To find the answer to the given statements, we need to calculate the reaction quotient (Q) for reaction (1)

 (2)

We need to remember that only gaseous compounds or liquids participate in the expression of K (equilibrium constant) or Q (reaction quotient).

The values of concentrations of the mixture are:

By introducing the above values into equation (2), we have:

Since Q > Kc, the reaction will shift in the direction of reactants to reduce the excess concentration of products and thus reach equilibrium.  

A value of Q = K means that the reaction is at equilibrium, and a value of Q < K means that the reaction will shift in the direction of products because of the excess of reactants.    

Therefore, the answer is the reaction will shift in the direction of reactants (option C).

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Rxn will shift toward reactant side of equation b/c Kc < Qc

Explanation:

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

i believe it is c.) 1575 j.