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.

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

$\text{Rate}=(0.0481)\times (1.48)^2(1.32)^1$

$\text{Rate}=0.14Ms^{-1}$

Therefore, the rate for a reaction will be $0.14Ms^{-1}$