Question

The rate constant for a second order reaction (second order in A) is k = 0.132 L/mol/s. What is the rate of the reaction when the concentration of A is 0.158 mol/L?

Answer 1

Answer on Question #43304, Chemistry, Other

Question:

The rate constant for a second order reaction (second order in A) is $k = 0.132 \, \text{L/mol/s}$. What is the rate of the reaction when the concentration of A is $0.158 \, \text{mol/L}$?

Solution:

Generally, for a reaction of the form

$$aA + bB + \dots \rightarrow \text{products}$$

the rate law will be

$$r = k[A]^a [B]^b \dots$$

In this expression, the $k$ is the reaction rate coefficient or rate constant, the exponents $a$ and $b$ are the reaction orders and depend on the reaction mechanism, [A] and [B] is molar concentrations of species A and B.

If rate of reaction depends on only one reactant concentration, rate law equation simplifies to $r = k[A]^a$.

We can use this equation:

$$r = 0.132 \, \text{L/mol} \cdot s \times (0.158 \, \text{mol/L})^2 = 0.00330 \, \text{mol/L} \cdot s = 3.30 \times 10^{-3} \, \text{mol/L} \cdot s$$

Answer:

The rate of the reaction is $3.3 \times 10^{-3} \, \text{mol/L} \cdot s$

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