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 : 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:

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:

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:

where,

k = rate constant =

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:

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

the answer is na (sodium)