Effect of Concentration on Reaction Rate

RESULTS AND DISCUSSION REPORT—EXPERIMENT 3 (CHEMICAL KINETICS)

CALCULATIONS

Effect of Concentration on Reaction Rate

[S2O32-]initand [H+]init for each run, knowing the original concentrations and volumes of [S2O32-], [H+], and water used.
[S2O32-]init= __(M[S2O32-])(V[S2O32-])__ [H+]init= _____(M[H+])(V[H+])____ V[S2O32-]+V[H+]+V[water] V[H+]+V[S2O32-]+V[water]

Run 1 [S2O32-]init= (0.15 M)(10 mL) (10+3+2)mL = 0.1 M
[H+]init= (3 M)(2 mL) (2+3+10)mL = 0.4 M

Run 2 [S2O32-]init= (0.15 M)(5 mL) (5+8+2)mL = 0.05 M
[H+]init= (3 M)(2 mL) (2+8+5)mL = 0.4 M

Run 3 [S2O32-]init= (0.15 M)(2.5 mL) (2.5+10.5+2)mL = 0.025 M
[H+]init= (3 M)(2 mL) (2+10.5+2.5)mL = 0.4 M

Run 4 [S2O32-]init= (0.15 M)(5 mL) (5+1+1.5)mL = 0.1 M
[H+]init= (3 M)(1.5 mL) (1.5+1+5)mL = 0.6 M

Run 5 [S2O32-]init= (0.15 M)(5 mL) (5+1.5+1)mL = 0.1 M

[H+]init= (3 M)(1 mL) (1+1.5+5)mL = 0.4 M

Run 6 [S2O32-]init= (0.15 M)(5 mL) (5+2+0.5)mL = 0.1 M
[H+]init= (3 M)(0.5 mL) (0.5+2+5)mL = 0.2 M

1/t. (Note: This can be taken as a measure of the initial rate of the reaction. Explain why.)
Run No. Time (sec) 1/Time (sec-1) OR Initial Rate
1 42.25 2.367 x 10-2
2 100.5 9.950 x 10-3
3 293.54 3.407 x 10-3
4 65.07 1.537 x 10-2
5 89.11 1.122 x 10-2
6 140.62 7.111 x 10-3

*The total volume used for the runs are the same and time is the only factor that varies. With all else held constant, we can derive the initial rate as the inverse of time.

Order of Reaction with respect to [S2O32-], using the initial rates method.

(Rate of run 2) / (Rate of run 1) = (k2[S2O32-]2X[H+]2Y) / (k1 [S2O32-]1X[H+]1Y)

1 / 42.25= k2[0.1 M]X[0.4 M]Y
1 / 100.5 k1[0.05 M]X[0.4 M]Y

2.3787 = 2x

ln