Problem 135
Question
When concentrations of the reactants is increased sixteen times, the rate becomes two times. The reaction is of (a) \(1 / 4\) order (b) fourth-order (c) third-order (d) \(1 / 8\) order
Step-by-Step Solution
Verified Answer
The reaction is of \(\frac{1}{4}\) order.
1Step 1: Understand the Problem
To determine the order of the reaction, we need to relate how the rate of reaction changes with the concentration of the reactants. The question tells us that when the concentration is increased by 16 times, the rate increases by 2 times.
2Step 2: Recall the Reaction Rate Law
The rate of a reaction can be expressed as: \[ ext{Rate} = k[ ext{Reactant}]^n\]where \(k\) is the rate constant, \([ ext{Reactant}]\) is the concentration of the reactant, and \(n\) is the order of the reaction.
3Step 3: Set Up the Equation
Given that the concentration is increased 16 times and the rate is increased 2 times, we can set up the equation:\[2 = 16^n\]This equation arises from setting the new rate equal to the old rate times the increase factor from concentration, both in terms of power \(n\).
4Step 4: Solve for the Order
Solve the equation for \(n\):\[2 = 16^n\]Express 16 as a power of 2: \[16 = 2^4\]Substitute this in to get:\[2 = (2^4)^n = 2^{4n}\]Since the bases are the same, equate the exponents:\[1 = 4n \]Therefore, solve for \(n\):\[n = \frac{1}{4}\]
5Step 5: Verify the Answer
Substituting \(n = \frac{1}{4}\) into the expression confirms that the rate indeed becomes 2 times when the reactant is increased 16 times. Thus, the reaction is of \(\frac{1}{4}\) order.
Key Concepts
Reaction Rate LawRate ConstantReactant ConcentrationExponent Solving
Reaction Rate Law
A fundamental concept in understanding chemical reactions is the reaction rate law. This law helps us predict how the rate of a reaction changes as the concentration of reactants changes. Essentially, the rate of a reaction can be expressed by the equation:
- \( \text{Rate} = k[\text{Reactant}]^n \)
Rate Constant
The rate constant, symbolized as \( k \), is a crucial component of the reaction rate law. It acts as a proportionality factor in the rate equation:
- \( \text{Rate} = k[\text{Reactant}]^n \)
Reactant Concentration
Reactant concentration is a key factor in the reaction rate law equation. It refers to how much of a reactant is present in a chemical reaction mixture at a given time. As seen in the formula:
- \( \text{Rate} = k[\text{Reactant}]^n \)
Exponent Solving
Being able to solve for the exponent in the reaction rate law is crucial for determining the reaction order. In our problem, one needs to solve the equation \( 2 = 16^n \) to find \( n \). To simplify this equation:
- First, express 16 as a power of 2: \( 16 = 2^4 \)
- Substitute into the equation: \( 2 = (2^4)^n = 2^{4n} \)
- Now that the bases are the same, equate the exponents: \( 1 = 4n \)
- Finally, resolve for \( n \): \( n = \frac{1}{4} \)
Other exercises in this chapter
Problem 131
The data given below is for the reaction of \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\) to form \(\mathrm{NOCl}\) at 295 \begin{tabular}{lll} \multicolumn{2}{c} { T
View solution Problem 134
Which of the following statements are correct about half-life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half-life peri
View solution Problem 137
For the reaction \(\mathrm{a} \mathrm{A} \longrightarrow \mathrm{xP}\) when \([\mathrm{A}]=2.2 \mathrm{mM}\) the rate was found to be \(2.4 \mathrm{~m} \mathrm{
View solution Problem 138
If the initial concentration of reactant in certain reaction is doubled, the half life period of the reaction is also doubled. The order of reaction is (a) zero
View solution