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

Answer:

m = 1

n = 1

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:

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

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

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