Question

In the oxidation of sulfur dioxide to sulfur trioxide, if 1.00 mol of SO2 and 1.00 mol of O2 are placed in a 1.00L flask at 1000K and at equilibrium, 0.925 mol of SO3 has been formed, what is the equilibrium constant K for this reaction?

A) K = 0.975
B) K = 0.925
C) K = 0.050
D) K = 1.075

If 1.00 mole of H2 and I2 are placed in a 0.5L flask at 425°C for the reaction H2(g) + I2(g) ⇌ 2HI(g) with Kc = 55.64, what are the concentrations of H2, I2, and HI when at equilibrium?
A) [H2] = 0.01 M, [I2] = 0.01 M, [HI] = 0.02 M
B) [H2] = 0.01 M, [I2] = 0.02 M, [HI] = 0.04 M
C) [H2] = 0.01 M, [I2] = 0.02 M, [HI] = 0.01 M
D) [H2] = 0.02 M, [I2] = 0.01 M, [HI] = 0.02 M

If the initial I2 concentration is 0.45 M for the reaction I2(g) ⇌ 2I(g) with K = 5.6 x 10^-12, what is the concentration of I and I2 at equilibrium?
A) [I2] = 0.45 M, [I] = 0.225 M
B) [I2] = 0.225 M, [I] = 0.45 M
C) [I2] = 0.225 M, [I] = 0.15 M
D) [I2] = 0.15 M, [I] = 0.225 M

Answer 1

The equilibrium constant for the given reaction is K = 0.050. Option C is correct.The concentrations of
`H_(2)`,
`I_(2)`, and HI when at equilibrium is [
`H_(2)`] = 0.01 M, [
`I_(2)`] = 0.02 M, [HI] = 0.01 M. Option C is correct. The concentration of I and
`I_(2)` at equilibrium is [
`I_(2)`] = 0.45 M, [I] = 0.225 M. Option A Is correct.

Step-by-step explanation:

1. To calculate the equilibrium constant, K, for the reaction 2SO₂(g) + O₂(g) = 2SO₃(g), we need to use the concentrations at equilibrium.

Given [SO₂] = 0.90 M, [O₂] = 0.35 M, and [SO₃] = 1.1 M,

we can substitute these values into the equilibrium constant expression: K = ([SO₃]² / ([SO₂]² * [O₂])).

Plugging in the values,

we get: K = (1.1²) / (0.90² * 0.35) = 0.050.

Therefore, the equilibrium constant for this reaction is K = 0.050. Option C is correct.

2. Equilibrium constant expression:

Kc = [HI]² / [
`H_(2)`][
`I_(2)`] = 55.64

Substituting the equilibrium concentrations into the expression:

55.64 = (2x)² / (1.00 - x)(1.00 - x)

Solving for x:

x = 0.01

Equilibrium Concentration (M)

`H_(2)` 1.00 - x = 1.00 - 0.01 = 0.99 M

`I_(2)` 1.00 - x = 1.00 - 0.01 = 0.99 M

HI 2x = 2(0.01) = 0.02 M

Hence, option C is correct.

3. The equilibrium constant expression for the reaction is:

`K_(c)` = [HI]² / [
`H_(2)`][
`I_(2)`]

Substituting the equilibrium concentrations into the expression, we get:

55.64 = (2x)² / (1.00 - x)(1.00 - x)

Solving for x, we get:

x = 0.02

Therefore, the equilibrium concentrations are:

[H2] = 1.00 - x = 0.98 M

[I2] = 1.00 - x = 0.98 M

[HI] = 2x = 0.04 M

Hence, option A is correct.