Question #286317

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.
1
Expert's answer
2022-01-18T03:01:03-0500

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


0.0180.008=k(0.100)x(0.300)yk(0.100)x(0.200)y\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=22.25 = (1.5)^y => y = 2

OR

0.0080.002=k(0.100)x(0.200)yk(0.100)x(0.100)y\dfrac{0.008}{0.002} = \dfrac{k(0.100)^x(0.200)^y}{k(0.100)^x(0.100)^y}

4=(2)y=>y=24 = (2)^y => y = 2


From 1, 4 and 5


0.0040.002=k(0.200)x(0.100)yk(0.100)x(0.100)y\dfrac{0.004}{0.002} = \dfrac{k(0.200)^x(0.100)^y}{k(0.100)^x(0.100)^y}

2=(2)x=>x=12 = (2)^x => x = 1

OR

0.0060.002=k(0.300)x(0.100)yk(0.100)x(0.100)y\dfrac{0.006}{0.002} = \dfrac{k(0.300)^x(0.100)^y}{k(0.100)^x(0.100)^y}

3=(3)x=>x=13 = (3)^x => x = 1


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

From experiment 1 the rate constant

k=0.002 mol dm3 s10.100 mol dm3 (0.100 mol dm3)2=2 dm6 mol2 s1k=\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}


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