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  2. The following data were obtained for the reaction : 
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The following data were obtained for the reaction : 

A+2B\rightarrow C

Experiment [A]/M [B]/M Initial rate of formation of C/M min-1
1 0.2 0.3 4.2\times 10^{-2}
2 0.1 0.1 6.0\times 10^{-3}
3 0.4 0.3 1.68\times 10^{-1}
4 0.1 0.4 2.40\times 10^{-2}

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of reaction. 

(c) Calculate the rate constant (k). 

 

 

 

 

 
 
 
 
 

Answers (1)
S Sumit

A+2B\rightarrow C

Supose order w.r.t A is \alpha and w.r.t B is \beta. 

Rate\: law[r=k[A]^\alpha [B]^\beta ]

\left ( Rate \right )exp_1,\: r_1=4.2\times 10^{-2}=k[0.2]^\alpha [0.3]^\beta\: \: \: \: \: \: \: \: \: \: ---(1)

                           r_2=6.0\times 10^{-3}=k[0.1]^\alpha [0.1]^\beta\: \: \: \: \: \: \: \: \: \: ---(2)

r_3=1.68\times 10^{-1}=k[0.4]^\alpha [0.3]^\beta\: \: \: \: \: \: \: \: \: \: ---(3)

r_4=2.40\times 10^{-2}=k[0.1]^\alpha [0.4]^\beta\: \: \: \: \: \: \: \: \: \: ---(4)

\frac{r_1}{r_3}, we\: \: \: get.

\frac{r_1}{r_3}=\frac{4.2\times10^{-2}}{1.68\times10^{-1}} =\frac{k[0.2]^\alpha[0.3]^\beta}{k[0.4]^\alpha[0.3]^\beta}

\Rightarrow \frac{25}{100}=\left [ \frac{1}{2} \right ]^\alpha

\Rightarrow \left ( \frac{1}{2} \right )^2=\left [ \frac{1}{2} \right ]^\alpha

\Rightarrow \alpha =2

similarly for \beta, \frac{r_2}{r_4}=\frac{6.0\times10^{-3}}{2.40\times10^{-2}}=\frac{k[0.1]^\alpha[0.1]^\beta}{k[0.1]^\alpha[0.4]^\beta}

\Rightarrow\frac{[0.1]^\beta}{[0.4]^\beta}=\frac{6.0\times10^{-3}}{2.40\times10^{-2}}=\frac{1}{4}\Rightarrow \beta =1      

(a) Hence, order w.r.t A is 2, w.r.t B is 1 

(b) Rate Law: 

r=k[A]^2[B]

overall order =2+1=3

(c)    k=\frac{r_1}{[0.2]^\alpha[0.3]^\beta}                       using eqn  (1)

             =\frac{4.2\times10^{-2}}{[0.2]^2[0.3]}=3.5mol^{-2}L^{-2}min^{-1}                 

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