Question

1. Assume that 5.00mL of .0500M Na2S2O3 is added to the reaction and that the 150.00 mL solution turns blue in 95 seconds. Calculate the reaction rate in mol I2/Ls. (This is also the rate in mol H2O2/Ls.

2. If a second experiment is done with half as much H2O2 as the experiment in question 1, and the solution turns blue in 180 seconds, what is the order of the reaction with respect to H2O2?

Thank you!!

Answer 1

1. 5 mL. of 0.05 M Na₂ S₂O₃ contains 5*0.05/ 1000 mole = 0.00025 mole Na₂ S₂O₃

0.00025 mole of Na₂ S₂O₃ reacts with 0.00025/2 mole of I₂ = 0.000125 mole of I₂

0.000125 mole I2 is dissolved in 150.00 mL of solution and it takes 95 seconds to complete the reaction.

The reaction rate is : 0.000125/(150*95) mole I2/ L s= 8.77 *10^-9 mole I₂/ L s

2. Suppose the reaction rate law : R = k * [H2O2]^x

Suppose in first case A mole H2O2/ L is consumed in 95 seconds

So, in first case, the rate of reaction is : A/ 95 mole/ Ls and rate law is : A/ 95 = k * [A]^x ( Equation no. 1)

In second case, the concentration of H2O2 is A/ 2 mol/ L and reaction takes 180 seconds

So, in second experiment rate law is: A/ 2*180 = k * [A/2]^x (Equation no. 2)

Dividing equation 1 by 2 we get,

2*180/95 = 2^x

or, 2² = 2^x

So, x = 2

The order of the reaction with respect to H2O2 is 2.