Question

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) 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) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of \(8 .\)

Short answer

For each case, the values of \(x\) are: (a) \(x = 0\) (b) \(x = 2\) (c) \(x = 3\)

Step-by-step solution

Case (a): No rate change when \([\mathrm{A}]_0\) is tripled.

In this case, the rate remains constant even after \([\mathrm{A}]_0\) is tripled. We can write this information as two equations: 1. rate = \(k[\mathrm{A}]^x\) 2. rate = \(k(3[\mathrm{A}])^x\) Since the rates are equal, we can set the two equations equal to each other: \(k[\mathrm{A}]^x = k(3[\mathrm{A}])^x\) Now, divide both sides by \(k[\mathrm{A}]^x\) to solve for \(x\): \(1 = 3^x\) Taking the logarithm of both sides: \(x \log{3} = \log{1}\) \(x = \frac{\log{1}}{\log{3}}\) Since \(\log{1} = 0\), the value of \(x\) is 0 for this case.

Case (b): Rate increases by a factor of 9 when \([\mathrm{A}]_0\) is tripled.

In this case, the rate increases by a factor of 9 when \([\mathrm{A}]_0\) is tripled. We can write this information as two equations: 1. rate1 = \(k[\mathrm{A}]^x\) 2. rate2 = \(k(3[\mathrm{A}])^x\) Since the second rate is 9 times the first rate, we can write it as: rate2 = 9 * rate1 Substituting the expressions for rate1 and rate2: \(k(3[\mathrm{A}])^x = 9k[\mathrm{A}]^x\) Divide both sides by \(k[\mathrm{A}]^x\): \(3^x = 9\) Taking the logarithm of both sides: \(x\log{3} = \log{9}\) Solving for \(x\): \(x = \frac{\log{9}}{\log{3}} = 2\) In this case, the value of \(x\) is 2.

Case (c): Rate increases by a factor of 8 when \([\mathrm{A}]_0\) is doubled.

In this case, the rate increases by a factor of 8 when \([\mathrm{A}]_0\) is doubled. We can write this information as two equations: 1. rate1 = \(k[\mathrm{A}]^x\) 2. rate2 = \(k(2[\mathrm{A}])^x\) Since the second rate is 8 times the first rate, we can write it as: rate2 = 8 * rate1 Substituting the expressions for rate1 and rate2: \(k(2[\mathrm{A}])^x = 8k[\mathrm{A}]^x\) Divide both sides by \(k[\mathrm{A}]^x\): \(2^x = 8\) Taking the logarithm of both sides: \(x\log{2} = \log{8}\) Solving for \(x\): \(x = \frac{\log{8}}{\log{2}} = 3\) In this case, the value of \(x\) is 3. In summary, we found the values of \(x\) for the three different cases: (a) x = 0 (b) x = 2 (c) x = 3

Key concepts

Chemical Kinetics

Understanding chemical kinetics is crucial for predicting how fast a reaction will occur. Fundamentally, it's the branch of physical chemistry that deals with the study of reaction rates and the steps involved in making a chemical reaction go from reactants to products. The rate of a reaction can be influenced by various factors, including reactant concentrations, temperature, and the presence of catalysts.

A key concept within kinetics is the rate law, which provides a mathematical relationship between the rate of a chemical reaction and the concentrations of reactants. In the case of the simple reaction \( \mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C} \), we express the rate law as \( \text{rate} = k[\mathrm{A}]^{x} \). Here, \( k \) is the rate constant, \( [\mathrm{A}] \) is the concentration of reactant A, and \( x \) indicates the reaction order with respect to A. By conducting experiments and observing how the rate changes when we alter \( [\mathrm{A}] \), we can determine the value of \( x \)—crucial for understanding how the reaction proceeds.

Rate Constant

The rate constant, denoted as \( k \), is a proportionality constant in the rate law equation that provides the rate of a chemical reaction at a given temperature. It's specific for each chemical reaction and is influenced by conditions such as temperature and the presence of a catalyst.

For the reaction \( \mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C} \), the rate law is defined by the expression \( \text{rate} = k[\mathrm{A}]^{x} \), making \( k \) a key factor in determining how quickly reactant A is converted into products B and C. Understanding the significance of \( k \) helps us to compare the innate speed of different reactions under the same conditions, and to predict how a reaction's rate will alter when the temperature or other conditions change.

Reaction Order

Reaction order refers to the power to which the concentration of a reactant is raised in the rate law equation. It indicates how the rate of reaction is affected by the concentration of that reactant. For example, in the provided exercise with the reaction \( \mathrm{A} \longrightarrow \mathrm{B} + \mathrm{C} \), the rate law is given by \( \text{rate} = k[\mathrm{A}]^{x} \), where \( x \) represents the reaction order for reactant A.

In the cases from the exercise, different scenarios unfold based on the value of \( x \) which gives us information about the reaction order. For instance, when \( x = 0 \), it shows that the rate is independent of the concentration of A, defining a zero-order reaction. When \( x = 2 \) or \( x = 3 \), these indicate second and third-order reactions, respectively, where the rate changes significantly with the concentration of A. Recognizing the reaction order is vital for controlling reaction rates in industrial processes as well as understanding mechanisms in chemical research.