Question

The reaction:so2cl2(g)---so2(g)+cl2(g) is the first order reaction .if the initial concentration of so2cl2 was 0.0248 mol/L and the rate constant is 2.2×10-⁵/s.what is the concentration of so2cl2 after two hours?

Answer 1

After two hours, the concentration of SO₂Cl₂ in the first-order reaction, described by ln([A]/[A]₀) = -kt, with an initial concentration of 0.0248 mol/L and a rate constant of 2.2 × 10^(-5) 1/s, is approximately 0.0212 mol/L.

The first-order reaction follows the integrated rate law:

ln([A]/[A]₀) = -kt

Where:

[A] is the concentration of the reactant at time t,

[A]₀ is the initial concentration,

k is the rate constant, and

t is the reaction time.

In this case, [A] is the concentration of SO₂Cl₂ at time t, and [A]₀ is the initial concentration of SO₂Cl₂.

We can rearrange the equation to solve for [A]:

`\[ [A] = [A]_0 \cdot e^(-kt) \]`

Given:

[A]₀ = 0.0248 mol/L

k = 2.2 × 10^(-5) 1/s

t = 2 hours = 7200 seconds

Now, plug in the values and solve:

`\[ [A] = 0.0248 \cdot e^{-(2.2 * 10^(-5) \cdot 7200)} \]`

[A] = 0.0248 * e^(-0.1584)

[A] ≈ 0.0248 * 0.8538

[A] ≈ 0.0212 mol/L

So, after two hours, the concentration of SO₂Cl₂ is approximately 0.0212 mol/L.