Question

1)Fill in the table given below, then drax a graph In (Na2S2O3] versus time(in seconds) by using excel. (20p) In Na 8,03] t(seconds) -1.02 0 - 1.20 900 -1.39 1800 -1.66 2880 -2.04 4500 In (Na2S0203) vs. Time(s) 0 1000 2000 3000 4000 5000 -0.5 -1 In (Na2S0203) -1.5 -2 -2.5 Time (s)

Answer 1

Part 3

The order is 1, i.e., the reaction is a first order reaction.

The graph is clearly a straight line with (-)ve slope. This is characteristic of a 1st order reaction, as derived below:

Let a be the initial concentration of Na2S2O3 and a-x be its concentration after a time t has elapsed. Then the rate of the reaction is clearly dx/dt. If k is the rate constant,

dx/dt = k(a-x) $\Rightarrow$ dx/a-x = k dt

Integrating LHS and RHS,

- ln(a-x) = kt +C, where C is the constant of integration.

This is of the form y = mx + c, where x = t and y = ln(a-x).

Hence, for any first order reaction, the graph between ln(a-x) and t will always be a straight line with (-)ve slope.

Part 5

Since this is a first order reaction, the rate of the reaction is directly proportional to the initial concentration of the reactant, so the rate will increase if [Na2S2O3] increases.

Part 6

From the equation above, when t = 0, x = 0, so C = -ln a.

Substituting the value of C,

- ln(a-x) = kt - ln a  $\Rightarrow$ k = (1/t) * ln(a/a-x)

In the given reaction, a = 1/2.78 = 0.36 M.

After a time t = 4500 s, a-x = 1/7.69 = 0.13 M.

Substituting these values in the equation, we get k = 2.263 * 10-4 /s.

Part 7

It is a linear relationship. Since the reaction obeys first order kinetics, the rate of the reaction R is proportional to [Na2S2O3], i.e., R = k * [Na2S2O3], where k is the rate constant (2.263 * 10-4 /s). This means that as the initial concentration of sodium thiosulfate is increased, the rate at which it is consumed in the reaction increases.