In a reaction, if the concentration of reactant A is halved, and the rate of rea
Practice Questions
Q1
In a reaction, if the concentration of reactant A is halved, and the rate of reaction decreases to one-fourth, what is the order of the reaction? (2020)
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Questions & Step-by-Step Solutions
In a reaction, if the concentration of reactant A is halved, and the rate of reaction decreases to one-fourth, what is the order of the reaction? (2020)
Step 1: Understand that the rate of a reaction depends on the concentration of the reactants.
Step 2: Identify that halving the concentration of reactant A means we change it from [A] to [A]/2.
Step 3: Note that the rate of reaction decreases to one-fourth when the concentration of A is halved.
Step 4: Use the formula for reaction rate: Rate = k[A]^n, where k is the rate constant and n is the order of the reaction.
Step 5: If the concentration is halved, we can express the new rate as Rate' = k([A]/2)^n.
Step 6: Simplify the new rate: Rate' = k[A]^n / 2^n.
Step 7: Since we know Rate' is one-fourth of the original rate, we can set up the equation: k[A]^n / 2^n = (1/4) * k[A]^n.
Step 8: Cancel k[A]^n from both sides of the equation, leading to 1/2^n = 1/4.
Step 9: Recognize that 1/4 can be rewritten as 1/2^2, so we have 1/2^n = 1/2^2.
Step 10: Conclude that n must equal 2, indicating that the reaction is second-order.
