Question

at a given temperature, 4.029 atm of cl2 and 2.495 atm of br2 are mixed and allowed to come to equilibrium. the equilibrium pressure of brcl is found to be 1.868 atm. calculate kp for the reaction at this temperature.

Answer 1

The equilibrium constant Kp for the reaction between Cl2 and Br2 to form BrCl can be determined by using the provided equilibrium pressures and the stoichiometry of the reaction. The pressures of reactants Cl2 and Br2 at equilibrium are calculated by assuming the change in pressure for each, based on the stoichiometry, and substituting these values into the Kp expression.

Step-by-step explanation:

The question involves finding the equilibrium constant Kp for a chemical reaction involving chlorine (Cl2) and bromine (Br2) to form bromine chloride (BrCl). According to the provided information, the initial pressures and the equilibrium pressure of one of the products (BrCl) are known. The reaction in question is:

Cl2(g) + Br2 (g) → 2BrCl(g)

To calculate Kp, you will use the equilibrium pressures in the expression for Kp which is:

Kp = (PBrCl2) / (PCl2 x PBr2)

Since we know the equilibrium pressure of BrCl (1.868 atm), and we have to assume that the changes in pressure for Cl2 and Br2 are equal to x. This assumption is based on the stoichiometry of the reaction. Thus, the partial pressures of Cl2 and Br2 at equilibrium will be (4.029 - x) atm and (2.495 - x) atm, respectively.

Then, we can establish the relationship between the pressure of BrCl (1.868 atm) and the changes in the pressures of Cl2 and Br2 by considering the stoichiometry of the reaction:

(1.868 atm)^2 = x2

Now, we can use the values of x to find the equilibrium pressures of Cl2 and Br2 and substitute them into the Kp expression to find the value of Kp.

Answer 2

The equilibrium constant
`\(K_p\)` for the reaction at this temperature is approximately 0.3476 (rounded to four decimal places).

To calculate the equilibrium constant Kp for the reaction
`\(Cl_2(g) + Br_2(g) \rightleftharpoons 2BrCl(g)\)`, you can use the equilibrium pressures of the reactants and products. The expression for Kp is given as:

`\[K_p = \frac{{[BrCl]^2}}{{[Cl_2][Br_2]}}\]`

We are given the following equilibrium pressures:

`\(P_(BrCl) = 1.868 \, \text{atm}\)`

`\(P_(Cl_2) = 4.029 \, \text{atm}\)`

PBr2 =2.495atm

Now, let's calculate Kp step by step:

1. Plug the given pressures into the expression for Kp:

`\[K_p = \frac{{(1.868 \, \text{atm})^2}}{{(4.029 \, \text{atm})(2.495 \, \text{atm})}}\]`

2. Calculate the numerator:

`\[Numerator = (1.868 \, \text{atm})^2 = 3.487024 \, \text{atm}^2\]`

3. Calculate the denominator:

`\[Denominator = (4.029 \, \text{atm})(2.495 \, \text{atm}) = 10.047975 \, \text{atm}^2\]`

4. Calculate
`\(K_p\)`by dividing the numerator by the denominator:

`\[K_p = \frac{{Numerator}}{{Denominator}} = \frac{{3.487024 \, \text{atm}^2}}{{10.047975 \, \text{atm}^2}}\]`

5. Simplify the expression:

`\[K_p \approx 0.3476\]`