Question

For the reaction: H2(g) + Br2(g) ⇌ 2 HBr(g), Kc = 7.5 × 102 at a certain temperature. If 2 mole each of H2 and Br2 are placed in 2-L flask, what is the concentration of H2 at equilibrium?

A) 0.96 B) 0.93 C) 1.86 D)0.04 E) 0.07

Answer 1

The concentration of H2 at equilibrium is approximately 0.073 M. "0.07" is the concentration of H2 at equilibrium. So, the correct option is "'E" 0.07".

Step-by-step explanation:

The given reaction is: H2(g) + Br2(g) ⇌ 2HBr(g), with a given equilibrium constant (Kc) of 7.5 × 102. We are given that 2 moles each of H2 and Br2 are placed in a 2-L flask.

To find the concentration of H2 at equilibrium, we need to use the equilibrium expression and solve for the concentration of H2 at equilibrium. Using the equilibrium expression, we can set up the equation: Kc = [HBr]2 / ([H2] * [Br2])

Since the mole ratio of H2 to Br2 is 1:1, if x represents the concentration of H2 at equilibrium, then the concentration of Br2 at equilibrium would also be x. Substituting the given values into the equilibrium expression:

7.5 × 102 = (2x)2 / (x * x)

7.5 × 102 = 4x2 / x2

7.5 × 102 = 4 x2 = 4 / 7.5 × 102

x2 = 0.005333333

x ≈ √(0.005333333)

x ≈ 0.073

Therefore, the concentration of H2 at equilibrium is approximately 0.073 M.

"0.07" is the concentration of H2 at equilibrium. So, the correct option is "'E" 0.07".

Answer 2

The equilibrium concentration is 0.04 M.

The equilibrium constant in reactions

The equilibrium constant, often denoted as is a numerical value that expresses the ratio of concentrations of products to reactants at equilibrium in a chemical reaction.

We have that;

Kc =
`[HBr]^2/[Br_2] [H_2]`

Molarity of hydrogen = 2 mol/2 L = 1 M

Molarity of Bromine = 2mol/2 L = 1 M

`7.5 *10^2` =
`x^2/(1 - x)^2`

`7.5 *10^2`(
`x^2` - 2x + 1) =
`x^2`

`7.5 *10^2`
`x^2` - 15 *
`10^2`x + 7.5 *
`10^2`=
`x^2`

x = 0.963 M (Because x can not be greater than the initial concentration)

The equilibrium concentration of hydrogen gas = 1 M - 0.963 M = 0.04 M