Course Work Example on Chemical Kinetics

1.Temperature

The increase in temperature increases the rate of a chemical reaction since it increases the kinetic energy of vibration of molecules thus breaking chemical bonds. The collision theory states that the higher the kinetic energy of vibrating molecules in a chemical reaction the lower the activation energy needed hence higher rates of a reaction taking place. Activation energy is the amount of kinetic energy in a chemical reaction needed to overcome the transition stage thus making molecules to break up and take part in a reaction.

An increase in temperature increases the activation energy. This means that more molecules will have a lot of kinetic energy to vibrate at very high rates. This, in turn, makes it easy for molecules to take part in chemical reactions. The red curve represents the higher level of energy gained by the molecules after the temperature is increased.

2.Pressure

The increase in pressure increases the rate of a chemical reaction. This is because higher pressure increases the chances of molecules coming in contact.

Changing pressure only affect those reactions involving gasses.

3. Reactant concentration

Increasing the amount of the reactants in a chemical reaction increases the rate of a reaction. This works especially in the case of fluid reactions (Frank-Kamenetskii,2015)

4.Using a catalyst

It lowers the activation energy hence increasing rates of a reaction.

A-Reactant

B- Activated complex

C-Product

The energy changes taking place in diagram 1 is that the reactants undergo a chemical reaction. When the activation energy is sufficient, the molecules undergo an exothermic chemic reaction. The products are thus at a lower energy due to losing of energy to the surroundings (Semenov,2013).

Diagram B is similar to diagram A. However, if a catalyst were to be added for diagram B, the activation energy would be lower than that of diagram A.

Role of catalyst in industry

1.Homogenous Catalyst-Catalysis happens when the catalyst is in the same phase as the reactants and products. Sulfuric acid, for example, is used as a catalyst in the industrial manufacture of propanone, phenol and bisphenol A. It is rarely used since its expensive in to separate it from the products(Boudart,2014)

2.Heterogeneous Catalyst- it is a different phase of the reactants and products. Industrial use includes for example use of iron as a catalyst in the manufacture of ammonia and use of platinum and rhodium as catalysts in the manufacture of nitric acid (Boudart, & Djega-Mariadassou,2014).

In the Haber process, pure iron is normally used a catalyst. Potassium hydroxide is added to it to act as promoter. It reduces the activation energy in the chemical reaction hence making the combination of hydrogen and nitrogen easier (Porter,2013).

Part 3; Following reaction rates

Disappearing cross; Across marked on paper is normally placed under a beaker containing one or of the chemical reactants which are transparent. When another reactant is added into the beaker, a reaction takes place thus, making the liquid opaque due to the formation of new products. The faster the cross disappears, the faster the chemical reaction.

Part 4; Rate equations

1.Initial rate - in chemical kinetics this is the rate of reaction at an instant at the start of a reaction.It is a differential value of the products and reactants and can be found by finding the negative slope of a curve of centration of a particular reactant versus the time taken.

2.Order of a Reaction - In chemical kinetics is the exponential value given to a concentration of a reactant in the rate of reaction. The order gives the value of the concentration of a reactant taking part during a rate of reaction (Semenov,2013). They vary depending on the concentration orders of various reactions.

Finding reaction order for experimental data

Question 1

A (g)+B (g)C (g)

Rate Law for the reaction; Rate (moldm-3 s-1) = k [A]x[B]y

Taking the ratio of the rates of experiments 3 and 1;

Experiment 3 / Experiment 1= k [A]3x[B]3y / k [A]1x[B]y1

= [0.6/0.3] y *[0.3/0.15]x

= [2]x [2] y

Ylog2=log 2

Y=1

Xlog2=log2

X=1

(a)Order of reaction of a=1

(b)Order of reaction of b=1

(c)Rate constant value is given by

0.40 (moldm-3 s-1) = k [0.40] [0.1]

K=0.1 moldm-3 s-1

Question 2

(a)order of (H2O2)

mlog [5/5] =log [1]

m=1

I-

-nlog [5/5] =log 1

n=-1

z log [10/5] =Log2

Z=2

(b)

Rate = [H O2 2] / t

Initial rate equation = k'[I] n0 [H2O2]m0 where;

[I ]0 = initial concentration of the iodide

[H2O2]0= initial concentration of the hydrogen peroxide

(c)Rate constant for the reaction

2.8=k [5.6]2[5.6] [2.8]

K=0.00569 moldm-3 s-1

(d) overall rate of reaction

=m+n+z=4

Works Cited

Boudart, M. and Djega-Mariadassou, G., 2014. Kinetics of heterogeneous catalytic reactions. Princeton University Press.

Semenov, N.N., 2013. Some problems of chemical kinetics and reactivity (Vol. 1). Elsevier.

Porter, G. ed., 2013. Progress in reaction kinetics (Vol. 3). Elsevier.

Boudart, M., 2014. Kinetics of Chemical Processes: Butterworth-Heinemann Series in Chemical Engineering. Elsevier.

Frank-Kamenetskii, D.A., 2015. Diffusion and heat exchange in chemical kinetics. Princeton University Press.