Question

Enter your answer in the provided box. Remember to enter your answer to the correct number of significant figures. For the reaction: A(g) + B(g) → AB(g) the rate is 0.23 mol/L·s, when [A]0 = [B]0 = 1.0 mol/L. If the reaction is first order in B and second order in A, what is the rate when [A]0 = 2.0 mol/L and [B]0 = 4.6 mol/L?

Answer 1

Answer : The rate law is 4.232 mol/L.s

Explanation :

First we have to calculate the value of rate constant.

According to the question, the expression for rate law will be:

$Rate=k[A]^2[B]$

where,

k = rate constant

As we are given :

Rate law = 0.23 mol/L.s

Initial concentration of A = 1.0 mol/L

Initial concentration of B = 1.0 mol/L

Now put all the given values in the above rate law expression, we get:

$0.23mol/L.s=k\times (1.0mol/L)^2\times (1.0mol/L)$

$k=0.23M^{-2}s^{-1}$

Now we have to calculate the rate law when initial concentration of A and B is 2.0 mol/L and 4.6 mol//L respectively.

The expression for rate law will be:

$Rate=k[A]^2[B]$

where,

k = rate constant = $0.23M^{-2}s^{-1}$

Rate law = ?

Initial concentration of A = 2.0 mol/L

Initial concentration of B = 4.6 mol/L

Now put all the given values in the above rate law expression, we get:

$Rate=(0.23M^{-2}s^{-1})\times (2.0mol/L)^2\times (4.6mol/L)$

$Rate=4.232mol/L.s$

Therefore, the rate law is 4.232 mol/L.s