Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.

The concentration of A remaining after 100 s is 0.135 mol L⁻¹ (or approximately 0.14 mol L⁻¹). — This is a first order reaction question. The order is determined from the units of rate constant k = 2.0 × 10⁻² s⁻¹. According to Table 3.3 in the context, for a first order reaction, the units of k are s⁻¹, which matches the given value. For first order reactions, the integrated rate law is applied to find the remaining concentration.

Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹.…

Question 31 of 37intermediate🔧 ApplyNumerical2 marks

Consider a certain reaction $A \rightarrow \text{Products}$ with $k = 2.0 \times 10^{-2} \text{s}^{-1}$ . Calculate the concentration of $A$ remaining after 100 s if the initial concentration of $A$ is 1.0 mol L $^{-1}$ .

Correct Answer

The concentration of A remaining after 100 s is 0.135 mol L⁻¹ (or approximately 0.14 mol L⁻¹).

Exercise: EXERCISES | Q: 3.24 | (Page 27)

For More Understanding

Explanation

This is a first order reaction question. The order is determined from the units of rate constant k = $2.0 \times 10$ ⁻² s⁻¹. According to Table 3.3 in the context, for a first order reaction, the units of k are s⁻¹, which matches the given value. For first order reactions, the integrated rate law is applied to find the remaining concentration.

Solution Steps

Step 1: Identify the order of reaction from the units of rate constant k. Since k = $2.0 \times 10$ ⁻² s⁻¹ has units of s⁻¹, from Table 3.3, this is a first order reaction.
Step 2: Write the integrated rate law for first order reaction: ln[A] = ln[A]₀ - kt or equivalently [A] = [A]₀ × e^(-kt)
Step 3: Substitute the given values: [A]₀ = 1.0 mol L⁻¹, k = $2.0 \times 10$ ⁻² s⁻¹, t = 100 s
Step 4: Calculate: [A] = 1.0 × e^(- $0.02 \times 100$ ) = 1.0 × e^(-2) = $1.0 \times 0.1353$ = 0.135 mol L⁻¹

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