Question

In the chemical reaction: Br2(g) + Cl2(g) ↔ 2 BrCl(g) KP = 0.150 If there is initially 0.500 atm of BrCl and nothing else in a container, what direction does the reaction move: a) Q > K and the reaction moves to the right b) Q > K and the reaction moves to the left c) Q < K and the reaction moves to the left d) Q < K and the reaction moves to the right

Answer 1

We know that

Kp = Product of partial pressures of products / product of partial pressures of reactants

For given reaction

Br2(g) + Cl2(g) ↔ 2 BrCl(g)

Q = p^2BrCl / pBr2 X pCl2   [Q= reaction quotient]

now as given that there is only BrCl in the container so the pressure of BrCl will be only pressure to be accounted

Hence

Q= (0.5)^2 = 0.25

Now as Q > Kc The reaction will move in backward direction or left side or reactant side s

so answer is

Q > K and the reaction moves to the left

Answer 2

The reaction will  B) move to the left since Q > K.

FURTHER EXPLANATION

The equilibrium constant, K, describes the reaction at equilibrium. It provides insight on whether the formation of the reactants or products are favored at equilibrium.

The value of the equilibrium constant is determined from the following expression:

$K \ = \frac{[products]}{[reactants]}$

It is simply the ratio of the concentration of the products and the concentration of reactants at equilibrium.

If the value of the equilibrium constant, K, is greater than 1, then, at equilibrium, the formation of products is favored. If K is less than 1, then the formation of reactants will be favored.

For systems, however, that are not yet at equilibrium, the direction of the reaction that will lead to equilibrium can be determined using another constant called the reaction quotient, Q. It is obtained using the similar expression as K but instead of using the equilibrium concentrations of the reactants and products, their initial concentrations are used.

The value of Q indicates whether the formation of products or reactants are favored so that equilibrium is achieved. If Q is greater than K, then the formation of reactants (or the reverse reaction) is favored. If Q is less than K, then the formation of products (forward reaction) is favored.

In this problem, the reversible equation provided is:

Br₂(g) + Cl₂(g) ⇄ 2 BrCl(g)            K= 0.150

To obtain the Q value, the initial concentration (in this case the partial pressure)  of the substances are plugged into the equilibrium expression. Since there are no  Br₂(g) and Cl₂(g) initially, only the partial pressure of BrCl will be used in the equation:

$Q = \frac{[BrCl]^2}{[Br_{2}][Cl_{2}]} \ or \ Q = \frac{P_{BrCl}^2}{P_{Br_{2}}P_{Cl_{2}}}\\\\Q = P_{BrCl}^2\\Q = (0.500 \ atm)^2\\\boxed {Q = 0.250}$

Comparing the Q vs. K,

Q = 0.250 and K = 0.150, hence,

$0.25 > 0.150\\\\\boxed {\boxed {Q > K, the \ reverse \ reaction \ will \ be \ favored; \ the \ reaction \ moves \ to \ the \ left}}$

LEARN MORE

1. Equilibrium  brainly.com/question/538664
2. Le Chatelier's Principle https://brainly.com/question/12983923

Keywords: Equilibrium Constant, Reaction Quotient