2H2S(g)---2H2(g) + S2 delta H=220KJ
k eq is 0.40
a 10.0L vessel containing 2.0 mol of H2S, 2.2 mol of H2, and 2.5 mol of S2
Calculate a trail k eq value to see whether the reaction is at equilibrium.
trail k eq =(h2)^2(S2)/(H2S)^2
= (0.22)^2(0.25)/(0.2)^2
= 0.3025 therefor keq < k eq
In which direction must the system move in order to reach equilibrium? why?
The equation must shift rigth because the numerator is not large enough. Since the numerator contains the product more product must be made in order to bring the ratio up to the value of 0.40
right
To determine the direction in which the system must move to reach equilibrium, we need to compare the calculated trial K_eq value to the actual K_eq value given in the problem.
The trial K_eq value is 0.3025, whereas the actual K_eq value is 0.40. Since the trial K_eq is less than the actual K_eq, it means that the system is currently not at equilibrium and needs to shift in a direction that increases the ratio of products to reactants.
In this case, the numerator of the K_eq expression represents the concentration of the products, while the denominator represents the concentration of the reactant. The trial K_eq value is smaller than the actual K_eq, which means that the concentration of products is currently lower than the concentration of the reactant.
To reach equilibrium, the system needs to shift in a direction that favors the formation of more products. This means that the system needs to move towards the right side of the equation, forming more H2 and S2 molecules while consuming H2S molecules.
Overall, the system needs to move in the forward direction (to the right) to reach equilibrium. This will increase the concentration of the products and bring the ratio closer to the actual K_eq value of 0.40.