Question

A zero-order reaction is 40% complete at the end of 50 minutes. What is the value id the rate constant with proper units? In many minutes will the reaction be 80% complete?

Answer 1

Suppose the reaction is

A => B

Rate of reaction

Rate of disappearance of A (-r_A)= Rate of formation of B (+r_B) = k ........where k is rate constant

-dC_A/dt = k

-dC_A= kt

integrate the above equation in the limits C_A0 to C_A at t= 0 to t

So the integration within the limits wil give

C_A0- C_A= -kt

(C_A0= initial concentartion in mol/L of A at t=0 mins, C_A= concentartion of A in mol/L at t=t mins)

Now 40% of reaction went to completion in 50 mins.

Means the 40% of A is reacted to produce B in 50 mins.

So,

C_A0- C_A = C_A0 - 0.60 C_A0 = 0.40 C_A0

And t= 50 mins

So, put this in above equation

0.40 C_A0 = -50k

k= - (0.40/50) C_A0

k= -8*10^-3 C_A0 mol/(L.min)

Now for 80% completion

C_A0-C_A = 0.80C_A0

Then,

0.80C_A0 = -kt

0.80C_A0 = - (-8*10^-3) C_{A0 *} t

t= 100 mins

So in 100 mins reacion will be 80% complete.