Question

Use this Initial Rate data to answer Questions 6 and 7. Consider the reaction:

xA + yB → products

The following Initial Rate kinetic data were obtained:

Trial [A] [B] Initial Rate (mol/L-s)
1 0.100 0.400 0.0904
2 0.200 0.800 0.362
3 0.400 1.200 1.08
4 0.200 0.400 0.181
5 0.100 0.800 0.181

The rate law for this reaction is: rate = k[A]m[B]n.
(6) What is m (the reaction order with respect to reactant A)?
(7) What is n (the reaction order with respect to reactant B)?

Answer 1

m = 1

n = 1

Step-by-step explanation:

The rate law is:

`r=k.[A]^(m) .[B]^(n)`

where,

r is the rate of the reaction

k is the rate constant

m is the order of reaction with respect to A

n is the order of reaction with respect to B

Let's consider trials 1 and 4. We know that [B]₁ = [B]₄ . The rate r₁/r₄ is:

`(r_(1))/(r_(4)) =(k.[A]_(1)^(m).[B]_(1)^(n)  )/(k.[A]_(4)^(m).[B]_(4)^(n)) \\(r_(1))/(r_(4)) =(([A]_(1))/([A]_(4)) )^(m) \\(0.0904M/s)/(0.181M/s)=((0.100M)/(0.200M))^(m) \\m=1`

Let's consider trial 1 and 5. We know that [A]₁ = [A]₅. The rate r₁/r₅ is:

`(r_(1))/(r_(5)) =(k.[A]_(1)^(m).[B]_(1)^(n)  )/(k.[A]_(5)^(m).[B]_(5)^(n)) \\(r_(1))/(r_(5)) =(([B]_(1))/([B]_(5)) )^(n) \\(0.0904M/s)/(0.181M/s)=((0.400M)/(0.800M))^(n) \\n=1`