Answer to Question #286317 in Physical Chemistry for Trevor

Question

Topic: Rate Equations

Question 1

Twi compounds, A and B , react in the following way:

A + 3B —> AB₃.

using the data below, derive the rate equation for this reaction and explain your reasoning.(4 marks)

b) Use the experimental data provided below to determine the rate constant. ( 3 marks)

data:

• Experiment 1 ,Initial (A) ( mol dm^-3) 0.100 , Initial (B) (mol dm³ ) 0.100 , Initial rate of formation of AB (mol dm^-3)0.002.
• Experiment 2, Initial (A) (mol dm^-3) 0.100 , Initial (B) ( mol dm^-3) 0.200, Initial rate of formation of AB₃ ( mol dm^-3) 0.008.
• Experiment 3 , Initial (A) (mol dm^-3) 0.100 , Initial (B) (mol dm^-3) 0.300, Initial rate of formation of AB₃ (mol dm^-3) 0.018.
• Experiment 4, Initial (A) ( mol dm^-3) 0.200 , Initial (B) ( mol dm^-3) 0.100 , Initial rate of formation of AB ( mol dm^-3) 0.004.
• Experiment 5, Initial (A) (mol dm^-3) 0.300 , Initial (B) (mol dm^-3) 0.100 , Initial rate of formation of AB (mol dm^-3) 0.006.

Answer 1

Let the rate be given by r = k[A]^x[B]^y

1) 0.002 = k(0.100)^x(0.100)^y

2) 0.008 = k(0.100)^x(0.200)^y

3) 0.018 = k(0.100)^x(0.300)^y

4) 0.004 = k(0.200)^x(0.100)^y

5) 0.006 = k(0.300)^x(0.100)^y

From 1, 2 and 3

$\dfrac{0.018}{0.008 } = \dfrac{k(0.100)^x(0.300)^y}{k(0.100)^x(0.200)^y}$

$2.25 = (1.5)^y => y = 2$

OR

$\dfrac{0.008}{0.002} = \dfrac{k(0.100)^x(0.200)^y}{k(0.100)^x(0.100)^y}$

$4 = (2)^y => y = 2$

From 1, 4 and 5

$\dfrac{0.004}{0.002} = \dfrac{k(0.200)^x(0.100)^y}{k(0.100)^x(0.100)^y}$

$2 = (2)^x => x = 1$

OR

$\dfrac{0.006}{0.002} = \dfrac{k(0.300)^x(0.100)^y}{k(0.100)^x(0.100)^y}$

$3 = (3)^x => x = 1$

The rate equation for this reaction r = k[A][B]²

From experiment 1 the rate constant

$k=\dfrac{0.002\ mol\ dm^{-3}\ s^{-1}}{0.100\ mol\ dm^{-3}\ *(0.100\ mol\ dm^{-3})^2}=2\ dm^6\ mol^{-2}\ s^{-1}$