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  • Help me solve this At 5188o C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s−1 when 5% had reacted and 0.5 Torr s−1 when 33% had reacted. The order of the reaction is :

At 5188o C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s−1 when 5% had reacted and 0.5 Torr s−1 when 33% had reacted. The order of the reaction is :

  • Option 1)

    0

  • Option 2)

    2

  • Option 3)

    3

  • Option 4)

    1

 

Answers (1)

best_answer

As we learned

nth order reaction -

The rates of the reaction is proportional to nth power of reactant

- wherein

Differential rate law

=\frac{dx}{dt}=k(a-x)^{n}

Integrated rate laws,

=\frac{1}{n-1}[(a-x)^{1-n} -a^{(1-n)}]=k_nt

a= initial, concentration of reactant at t=0 sec

x= concentration of product formed at t= tsec

t_\frac{1}{2}=\frac{1}{(n-1)(a^{n-1})(k_n)}[2^{n-1}-1]

Formulae for all the order except n=1

 

 

 Let rate law be:

r=K\left [ acetaldehyde \right ]^{n}

\therefore 1= K\left ( 0.95\: P_{o} \right )^{n}

0.5=K\left ( 0.67\: P_{o})^{n} \right

So\: 2= \left ( \frac{0.95}{0.67} \right )\Rightarrow 2= \left ( 1.41 \right )^{n}\therefore n;2

 


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