Question

Ammonium cyanate, NH₄NCO, rearranges in water to give us urea., (NH₂)₂ CO: NH₄NCO (AQ) to ((NH₂))₂CO (AQ). The rate for this equation for the process is "rate =k (NH₄NCO)²," where k =0.113 L/mol per min. If the original concentration in the solution is 0.229mol/L, how long will it take for the concentration to decrease to 0.180 mol/L?

Answer 1

Given:

Rate law:

Rate = k [ NH4NCO2]²

k is rate constant and its value is 0.113 L / ( mol min)

Initial concentration is 0.229 mol / L

Final concentration 0.180 mol/L

Solution:

From the rate law we can say that the overall order of this reaction is two. To calculate the time required for this reaction we have to use integrated rate equation of second law.

Equation is :

1 /[A]_t= kt + 1/[A]₀

Where [A]_t is concentration of reactant at time ‘t’, [A]₀ is original concentration of reactant.

Lets plug given values,

1 / 0.180 = (0.113 t ) + 1/0.229

t = 10.5 min

Answer: Reaction would take 10.5 min long