Effect of Concentration on Reaction Rate

Topics: Reaction rate, Chemical kinetics, Rate equation Pages: 5 (812 words) Published: January 2, 2013
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
142.252.367 x 10-2
2100.59.950 x 10-3
3293.543.407 x 10-3
465.071.537 x 10-2
589.111.122 x 10-2
6140.627.111 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 (2.3787) = x ln (2)

x = ln (2.3787) / ln (2)= 1.25≈ 1

Order of Reaction with respect to [H+], using the initial rates method.

(Rate of run 6) / (Rate of run 5) = (k6[S2O32-]6X[H+]6Y) / (k5[S2O32-]5X[H+]5Y)

1 / 140.62 = k6[0.1 M]X[0.2 M]Y
1 / 89.11k5 [0.1 M]X[0.4 M]Y

0.6337 = 0.5Y

ln (0.6337) = y ln (0.5)

y = ln (0.6337) / ln (0.5)= 0.66 ≈ 0

Rate Law for the reaction

rate = k [S2O32-]1[H+]0
k = rate/ [S2O32-]1
k = 2.367 x 10-2s-1/ 0.1
k = 0.2367 s-1
rate = 0.2367 s-1[S2O32-]

Effect of Temperature on Reaction Rate

[S2O32-]init and [H+]initfor run 3.
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

1/t (s-1) and 1/T (K-1) (Note: In this particular case, 1/t is a measure of the rate constant, k. Explain why.) Temp (K)1/Temp (K-1)Run Time (s)1/Time (s-1)
280.653.563 x 10-311178.952 x 10-4
303.153.299 x 10-3293.543.407 x 10-3
343.152.914 x 10-333.742.964 x 10-2

*It is assumed that the order of the reaction with respect to the reactants is first overall; hence their concentration is deemed constant with time. As a result, rate constant is equal to the rate.

Eaof the reaction, using the slope of the graph of ln(1/t) as ordinate versus 1/T as abscissa. ln k = ln A – (Ea/R)( 1/T)
y = b + mx
ln k, s-1 (y)1/T, K-1 (x)
ln (8.952 x 10-4)3.563 x 10-3
ln (3.407 x 10-3)3.299 x 10-3
ln (2.964 x 10-2)2.914 x 10-3

Through linear regression, we get

m = -5409.51 K-1
b = 12.2215 s-1

m = -Ea/ R

Ea= -(R)(m)= -(8.314 J/mol K)(-5408.82 K-1) = 44975 J/mol
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