For a reaction, $ 2A + B \to \text{products} $, if concentration of B is kept constant and concentration of A is doubled, then rate of reaction is
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In reactions, when the concentration of a reactant is changed, the rate of reaction is affected by the order of the reaction with respect to that reactant. For a second-order reaction, doubling the concentration of a reactant will quadruple the rate.
To determine how the rate of reaction changes when the concentration of A is doubled, we start by considering the rate law for the reaction $2A + B \to \text{products}$. Assuming a general rate law of the form \(r = k[A]^m[B]^n\), where \(r\) is the rate of reaction, \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders with respect to \([A]\) and \([B]\) respectively.
Since the concentration of B is kept constant, our focus is on the relationship between \([A]\) and the rate of reaction. From the stoichiometry of the reaction, it's typical to assume that the reaction order \(m\) for \([A]\) is equal to its coefficient in the balanced equation, which is 2. Hence, the rate law can be approximated as \(r = k[A]^2[B]^n\), and simplifying with constant \([B]\) into \(r \sim [A]^2\).
Now, if the concentration of A is doubled, the new rate of reaction \(r'\) will be:
\(r' = k(2[A])^2[B]^n = 4k[A]^2[B]^n = 4r\).
This indicates that doubling the concentration of A causes the rate of reaction to increase by a factor of four, or the rate is quadrupled.
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Approach Solution -2
To determine how the rate of reaction changes when the concentration of A is doubled, we examine the rate law for the reaction given by:
Rate = $k[A]^m[B]^n$
where k is the rate constant, [A] and [B] are the concentrations of A and B, and m and n are the reaction orders with respect to A and B, respectively.
Given the reaction: 2A + B → products, and if the concentration of B is kept constant while the concentration of A is doubled, we need to evaluate how this affects the rate assuming the rate law predominantly depends on [A].
Let's assume that through experimentation, it is determined that the reaction is second order with respect to A (m = 2) and first order with respect to B (n = 1). Thus, the rate law can be expressed as:
Rate = $k[A]^2[B]$
Now, if [A] is doubled, the new concentration [A'] = 2[A]. Substitute this into the rate law:
New Rate = $k(2[A])^2[B] = k × 4[A]^2[B]$
This simplifies to:
New Rate = $4 × (k[A]^2[B])$
Comparing the New Rate with the original rate (k[A]^2[B]), we see that the rate of reaction is quadrupled.
Therefore, when the concentration of A is doubled, the rate of reaction increases fourfold, or is quadrupled.
