Question

You perform a series of experiments for the reaction \(\mathrm{A} \rightarrow 2 \mathrm{~B}\) and find that the rate law has the form, rate \(=k[\mathrm{~A}]^{x} .\) Determine the value of \(x\) in each of the following cases: (a) The rate increases by a factor of \(6.25,\) when \([\mathrm{A}]_{0}\) is increased by a factor of \(2.5 .(\mathbf{b})\) There is no rate change when \([\mathrm{A}]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2,\) when \([\mathrm{A}]_{0}\) is cut in half.

Short answer

In summary, the values of x for each case are: (a) x ≈ 2, (b) x = 0, and (c) x = 1.

Step-by-step solution

Case (a)

We are given that the rate increases by a factor of 6.25 when the initial concentration of A, [A]₀, is increased by a factor of 2.5. Let's denote the initial rate as r₁ and the new rate as r₂. We have: \(r_1 = k[A_{1}]^{x}\) \(r_2 = k[A_{2}]^{x}\) Where r₁ = initial rate, r₂ = new rate, A₁ = initial concentration of A and A₂ = new concentration of A. Now, we know that r₂ = 6.25r₁ and A₂ = 2.5A₁. We can substitute these into the equations and solve for x: \(6.25r_1 = k(2.5A_1)^x\) Divide by r₁: \(6.25 = (2.5)^x\) Now, take the logarithm on both sides: \(\log{6.25} = x\log{2.5}\) Then solve for x: \(x = \frac{\log{6.25}}{\log{2.5}} \approx 2\) So, the value of x in case (a) is approximately 2.

Case (b)

We are given that there is no rate change when [A]₀ is increased by a factor of 4. This means r₂ = r₁ or the rate remains the same. And A₂ = 4A₁. We can substitute these into the equations and solve for x: \(r_1 = k[A_{1}]^{x}\) \(r_1 = k[4A_{1}]^{x}\) Now, since r₁ = r₂, we can divide the second equation by the first equation: \(1 = (4)^x\) Taking the logarithm on both sides, we have: \(0 = x\log{4}\) Then solve for x: \(x = \frac{0}{\log{4}} = 0\) So, the value of x in case (b) is 0.

Case (c)

We are given that the rate decreases by a factor of 1/2 when [A]₀ is cut in half. So, r₂ = (1/2)r₁, and A₂ = (1/2)A₁. We can substitute these into the equations and solve for x: \(\frac{1}{2}r_1 = k(\frac{1}{2}A_1)^x\) Divide by r₁: \(\frac{1}{2} = (\frac{1}{2})^x\) Taking the logarithm on both sides: \(\log{\frac{1}{2}} = x\log{\frac{1}{2}}\) Then solve for x: \(x = \frac{\log{\frac{1}{2}}}{\log{\frac{1}{2}}} = 1\) So, the value of x in case (c) is 1. In summary, the values of x for each case are: (a) x ≈ 2, (b) x = 0, and (c) x = 1.

Key concepts

Rate Equation

A rate equation, also known as a rate law, describes how the rate of a chemical reaction depends on the concentration of reactants. The general form of a rate equation is given by: \[ \text{rate} = k [\text{A}]^x [\text{B}]^y \cdots \] where:

• \(k\) is the rate constant, a proportionality factor.
• \([\text{A}]\), \([\text{B}]\) are the concentrations of the reactants.
• \(x\), \(y\) are the reaction orders with respect to reactants A and B.

The rate equation provides critical information about how different reactants influence the speed of the reaction. When given experimental data, you can derive the exponents like \(x\) and \(y\) by observing how changes in concentration affect the reaction rate.

Reaction Order

Reaction order refers to the exponents in the rate equation and indicates how the reaction rate is affected by the concentration of its reactants. It is the sum of all powers to which the concentration terms are raised in the rate equation. For a single-reactant reaction, it’s simply the exponent attached to the concentration of that reactant.

• If a reaction is first-order with respect to a reactant \(\text{A}\), it means that the rate changes linearly with the change in concentration of \(\text{A}\).
• Second-order indicates that the rate is proportional to the square of the concentration of the reactant.
• Zero-order indicates the rate is independent of the concentration of the reactant.

In our example, different conditions provided different reaction orders for A: approximately 2 (second-order), 0 (zero-order), and 1 (first-order). Reaction order is critical in understanding how modifications in the reaction system or conditions will affect the overall reaction rate.

Concentration Effect

The concentration effect in a chemical reaction refers to how changes in the concentration of reactants affect reaction speed. This effect is directly captured in the reaction order and rate equation:

• An increase in concentration can speed up the reaction if the reaction order is positive.
• If the reaction order is zero, changes in concentration do not affect the rate.
• A negative reaction order, although rare, would mean an increase in concentration actually slows the reaction down.

For instance, if changing the concentration of a reactant does not change the rate, the concentration effect is zero, as seen in Case (b). Understanding how concentration affects rate is essential, especially in industrial applications where reaction speed is crucial. These insights enable chemists to manipulate reaction conditions to optimize processes for faster or more controlled reactions.