Chemical kinetics

Started by Pranjal Singh, January 11, 2022, 10:19:20 PM

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Pranjal Singh

Here it is
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chenbeier

The solution is answer 1/3* 10^4 sec. Its not on the list.

The reason its a reaction second order.

B(t) = 1/(k*t + 1/B0)

1,5 m = 1/(10^-4*t+1/3m))

t= 1/3*10^4 sec

chenbeier

If for compound  A is locked for

Then
A(t) = 1/(k*t + 1/A0)

0,5 m = 1/(10^-4*t+1/2m))

t= 3/2*10^4 sec is the solution. Answer B)

uma

Quote from: Pranjal Singh on January 11, 2022, 10:19:20 PM
Here it is
[attachment id=0 msg=3171]
The overall order of the reaction is second order and there is no equation to calculate half life if both reactants are different. However we can calculate using the concept that of pseudo first order and then order with respect to B is first order and this will make half life = ln2/k
k in question is 10^-4 .

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