The equilibrium reaction below has the kc = 3.93. if the volume of the system at equilibrium is increased from 2.00 liters to 8.00 liters, how and for what reason will the equilibrium shift? be sure to calculate the value of the reaction quotient, q, and use this to confirm your answer.
co (g) + 3h2(g) -ch4(g) + h2o (g)

Q = 62.9

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

Explanation:

Let's consider the following reaction.

CO(g) + 3 H₂(g) ⇄ CH₄(g) + H₂O(g)

The equilibrium constant (Kc) is:

If the volume is multiplied by 4 (2.00 L → 8.00 L), the concentrations will be divided by 4. The reaction quotient (Q) is:

Q > Kc, so the reaction will proceed to the left so that Q takes the value of Kc.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%