Question

For the reaction, A(g) + B(g) => AB(g), the rate is 0.385 mol/L.s when the initial concentrations of both A and B are 2.00 mol/L. If the reaction is second order in A and first order in B, what is the rate when the initial concentration of A = 1.48 mol/L and that of B = 1.32 mol/L. Give your answer to 2 decimal places

Answer 1

Answer : The rate for a reaction will be
`0.14Ms^(-1)`

Explanation :

The balanced equations will be:

`A(g)+B(g)\rightarrow AB(g)`

In this reaction,
`A` and
`B` are the reactants.

The rate law expression for the reaction is:

`\text{Rate}=k[A]^2[B]^1`

or,

`\text{Rate}=k[A]^2[B]`

Now, calculating the value of 'k' by using any expression.

`\text{Rate}=k[A]^2[B]`

`0.385=k(2.00)^2(2.00)`

`k=0.0481M^(-2)s^(-1)`

Now we have to calculate the initial rate for a reaction that starts with 1.48 M of reagent A and 1.32 M of reagents B.

`\text{Rate}=k[A]^2[B]^0[C]^1`

`\text{Rate}=(0.0481)* (1.48)^2(1.32)^1`

`\text{Rate}=0.14Ms^(-1)`

Therefore, the rate for a reaction will be
`0.14Ms^(-1)`