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Determine The Order of A Reaction-First Order or Second Order and Calculate Its Rate Constant

If we have a problem that we must determine the rate law of the reaction and we just given by a particular data, the easy step to solve the problem is by taking a plot of ln[reactant] versus time and 1/[reactant] versus time too. If we have a straight line between one of the two of the plots, thus the plot with the straight line is determined the order of the reaction. If the straight line is the plot which ln[reactant] versus time thus the reaction is first order, and the reverse is the second order. Take a look to the sample below:

A student has studied the decomposition of hydrogen peroxide, and he obtained the following data at a given temperature

• Is this reaction first order or second order?
• Determine the rate law and the integrated law
• What is the value of the rate constant for the reaction
• What is the half-life for the reaction under the condition of the experiment

To find whether the rate law of this reaction is fisrt order or second order, we should have to make a plot of ln[H2O2] versus time and 1/[H2O2] versus time. If the first graph is a straight line thus the reaction should be fisrt order. If the second graph is a straight line thus the reaction is second order. The data necessary to make these plots are as follows:

the resulting plots are shown in the figures below:

from the figures above we know that ln[H2O2] versus time is a straight line but 1/[H2O2] versus time is not a straight line. Thus the decomposition of H2O2 is a fisrt order reaction and we can write the rate law as:

rate = k[H2O2]

and the integrated rate law as

ln[H2O2] = -kt + ln[H2O2]o

Since the reaction is first order, the slope of the line equals -k where

slope = Change in y / cahnge in x
slope = -2.995 - (-2.501) / (3600-3000)
slope = -8.233×10^-4 s^-

k = -(slope) = 8.233×10^-4 s^-

The half life of the fisrt order reaction difined as

substituting the value of k we have

t1/2 = 0.693 / 8.233×10^-4 s^-
t1/2 = 841.735 s

Add comment May 30th, 2008

Calculate Actual Mass of Methyl Salicylate

Chemistry Tutor Site Problem

Methyl salicylate(oil of wintergreen) is prepared by heating salicylic acid(C7H6O3) with methanol (CH3OH)? In an experiment, 6.252g of salicylic acid is reacted with an excess of methanol. How many grams of oil of wintergreen is produced if the percent yield of the reaction is 68.63%?

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When you solve the problem using stoichiometry calculation, it means that you calculate the theoretical yield. The percent yield of the reaction gives you the actual yield of the product. To find the mass of the methyl salicylate we need first to calculate the molar mass of each species are:

Mr C7H6O3 = 138.123 g/mol
Mr C8H8O3 = 152.1494 g/mol
Mr CH3OH = 32.04 g/mol

We have 6.252g of salicylic acid to be reacted with excess of methanol thus the moles of salicylic acid are

moles salicylic acid = 6.252 g / 138.123 g/mol = 0.045 mol

the balanced reaction is :

C7H6O3 + CH3OH –> C8H8O3 + H2O

because the mole ration between C7H6O3 and C8H8O3 is 1:1 thus 0.045 mol of C7H6O3 will theoretically produce 0.045 mol of C8H8O3 too, and the mass of C8H8O3:

mass C8H8O3 = 0.045 mol x 152.1494 g/mol = 6.887 grams

remember that this is the theoretically mass of C8H8O3, because the yield of the reaction is 68.63% so the actual mass of C8H8O3 tha has been produced is:

actual mass C8H8O3 = 68.63% x 6.887 grams = 4.726 gr

see also
Stoichiometry Calculation of theoretical and percent yield of aspirin

Add comment May 28th, 2008

How To Find Formula for The Half-Life of Second Order Reaction

In this article we try to find the general formula for the half-life of fecond order reaction. take a look into a general reaction that involve a single reactant and this reaction is a second order in the reactant (A), then the reaction can be writen as

aA -> products

The rate law for the reaction is

Rate = k[A]²

and the integrated second order reaction can be write as

The important meaning of the equation above is a plot of 1/[A] versus t will produce a straight line with the slope of k

when t1/2 is reached the concentration of [A] is equal to 1/2[A] and the equation above can be writen as

1/ (1/2[A]o) = kt1/2 + 1/[A]o
2/[A]o - 1/[A]o = kt1/2
1/[A]o = kt1/2

solving for t1/2 will give the expression for the half life of the second order reaction

This is the general equation to find half life of second order reaction.

Add comment May 28th, 2008

How To Find Formula for The Half-Life First Order Reaction

We can get general formula for the half-life of a first order reaction that erived from the integrated rate law for the general reaction:

aA -> products

The rate law is

Rate = k[A]

and the integrated fisrt order rate law is

ln[A] = -kt + ln [A]o

the important thing of this equation is that this is the y = mx + c where plot of y versus x is a straight line with the slope is m and the intercept c. Where y is ln[A], x is t, m= -k, and the c = ln[A]o. The plot of ln[A] versus t is the straight line it means that the reaction is first order when the plot is not a straight line the reaction is not first order. This is how we can test whether a reaction is the first order reaction or not.

or we can be expressed in term of a ratio as:

when t1/2 we have the concentration of [A] = 1/2 [A]o then the integrated formula becomes

ln { [A]o / (1/2 [A]o) } = k.t1/2
ln(2) = k.t1/2

because ln (2) is 0.693 then substituting this value to the equation above gives

the equation above is the formula to find half life of first order reaction. Just remember that for the fisrt order reaction the half life does not depend on the concentration

Add comment May 28th, 2008