Answer:
8.409 mol/L.s
Explanation:
A(g) + B(g) → AB(g)
It is given that the reaction is first order in B and second order in A
R = k [A]² [B]
where R = rate of the reaction
k = rate constant
[A] = concentration of reactant A at any time
[B] = concentration of reactant B at any time
At the start of the reaction,
rate is 0.25 mol/L·s, when [A]₀ = [B]₀ = 1.0 mol/L
Substituting these into the rate equation, we solve for the rate constant
0.25 = k (1²) (1)
k = 0.25 L²/mol².s
The rate when [A]₀ = 3.1 mol/L and [B]₀ = 3.5 mol/L
R = 0.25 [A]² [B]
R = 0.25 × (3.1)² × (3.5)
R = 8.40875 = 8.409 mol/L.s
Hope this Helps!!
