Question

What is the difference between a first order and a second order reaction?

a. For a first order reaction, plotting the natural logarithm of A versus time always gives a straight line with a slope equal to -k For a second order reaction a plot of 1 / A versus time, will produce a straight line with a slope equal to k

b. For a first order reaction, plotting the natural logarithm of A versus time always gives a straight line with a slope equal to -k For a second order reaction a plot of A versus time, will produce a straight line with a slope equal to k

c. For a first order reaction, plotting the natural logarithm of A versus time always gives a straight line with a slope equal to k For a second order reaction a plot of 1 / A versus time, will produce a straight line with a slope equal to -k

d. For a first order reaction, plotting the natural logarithm of A versus time always gives a straight line with a slope equal to -k For a second order reaction a plot of A versus time, will produce a straight line with a slope equal to -k

Answer 1

We know that the Integrated rate law equation, for the 1st order is,

ln [A]t = -kt + ln[A0] ..............(1)

where [A]t = concentration of A at time t.

[A0] = initial concentration

When we plot the graph ln [A]t vs t, we will get a straight line as this equation follows straight line equation i.e y =mx + c ......... (2)

where m = slope of the straight line.

on comparing equation (1) and (2), we get.

Equation (1) follows the straight-line equation.

and slope = -k for 1st order reaction.

Similarly, For the 2nd order,

we have an Integrated rate law equation,

1/ln [A]t = kt + 1/ln[A0] ..............(3)

on comparing equation (2) and (3), we get.

Equation (3) follows the straight-line equation.

and slope = k for 2nd order reaction.

Hence, For a first-order reaction, plotting the natural logarithm of A versus time always gives a straight line with a slope equal to -k For a second-order reaction a plot of 1 / A versus time, will produce a straight line with a slope equal to k.

Hence, Option (a) is the correct answer, and rest are incorrect.

Hope this helps!