For the reaction, A(g) + B(g) => AB(g), the rate is 0.625 mol/L. s when the initial concentrations of both A and B are 2.00 mol/L. If the reaction is first order in A and first order in B, what is the rate when the initial concentration of A = 4.41 mol/L and that of B = 2.72 mol/L. Give answer to 2 decimal places.

The rate of concentration is 1.87 mol L⁻¹s⁻¹

Explanation:

First order reaction: First order reaction is only depends on the value of the concentration of only one reactant.

A(g)+B(g)⇒AB(g)

rate of reaction = k [A][B]   [ since A and B are first order reaction]

k= rate constant

[A] = concentration of A

[B] = concentration of B

Given that rate = 0.625 mol/L.s

[A] = 2 mol/L   and   [B] = 2 mol/L

Therefore,

rate of reaction = k [A][B]

⇒0.625= k ×2×2

The value of k changes only when the temperature changes.

Now [A]= 4.41 mol/L    and   [B]=  2.72 mol/L

Rate of reaction = K[A][B]

 ≈1.87 mol L⁻¹s⁻¹

The rate of concentration is 1.87 mol L⁻¹s⁻¹

the answer is c i believe