Question

For a reaction that is either zero, first, or second order, which reaction order would leave the most reactant remaining after the reaction has occurred for an hour? Assume k does not change.

Answer 1

Zero = C vs t;

For zero order, there is no dependency of concentrations:

dC/dt = k*C^0

dC/dt = k

When developed:

C = C0 + kt

if x axis is "time" then the slope is "k", and y-intercept is initial concentration C0. y-axis if C (concentration)

First = ln(C) vs. t

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

if x axis is "time" then the slope is "-k", and y-intercept is initial concentration C0. y-axis if ln(C) (natural logarithm of concentration)

Second = 1/C vs. t

For Second order

dC/dt = k*C^2

When developed:

dC/C^2 = k*dt

1/C= 1/C0 + kt

Then...

assume C0 = 1 and C = x; for k = 0.01 and t = 10

the reactions:

C = C0 + kt

ln(C) = ln(C0) - kt

1/C= 1/C0 + kt

substitute

C = 1 - 0.01 *10 = 0.90 --> 90%

ln(C) = ln(C0) - kt --> ln(C) = ln(1) - 0.01*10 = -0.1 --> C = exp(-0.1) = 0.904 = 90.4%

1/C= 1/C0 + kt --> 1/(C) = 1/1 + 0.01*10 = C = 1.1^-1 = 0.909 --> 90.9%

Clearly;

Second order is going to leave a higher amount of reactant