S.No. |
A(M) |
B(M) |
Initial rate (M) |
1. |
1.00 |
1.00 |
1.2 x 10–2 |
2. |
1.00 |
2.00 |
4.8 x 10–2 |
3. |
1.00 |
4.00 |
1.9 x 10–1 |
4. |
4.00 |
1.00 |
4.9 x 10–2 |
Assuming that rate law can be written as
Determine the value of
The decomposition of H2O2 in basic solution is first order in H2O2.
2H2O2(aq) → 2H2O2 (l) x O2(g)
the rate constant is 1.6 x 10–5 s–1 at 25°C and initial concentration of H2O2 is 0.20 M.
(a) What is the concentration of H2O2 after 2 hrs.
(b) How long will it take for H2O2 concentration to drop to 0.08 M.
(c) How long will it take for 90% of H2O2 to decompose?
t in seconds |
0 |
900 |
1800 |
Cone. of A |
50.8 |
19.7 |
7.62 |
Prove that the reaction is of first order of A to decompose to one-half.
t/sec | 1242 sec | 2745 sec | 4546 sec |
At Conc. | -27.80ml | -29.70ml | -31.81ml |
Here [ A]0 is proportional to the alkali consumed between t = &&& and t = 339 sec, i.e., [A]0 = (39.81 mL – 26.34 mL) = 13.47 mL [A]t is proportional to the alkali consumed at &&& minus the alkali consmed at the specified time. Thus,
[A], at 1242 sec = 39.81 mL – 27.80 mL = 12.01 mL
[A], at 2745 sec = 39.81 mL – 29.70 mL = 10.11 mL
[A], at 4546 sec = 39.81 mL – 31.81 mL = 8.00 mL
Substituting the values of t, [A]0 and [A]t in the first-order rate equation
we get the following values of k
(i)
(ii)
(iii)
Since the three values of k derived from first-order reaction are close to being identical, hydrolysis of methyl acetate is a first order reaction.
t (in mitt) |
0 |
135 |
339 |
683 |
1680 |
C (mol L–1) |
2.08 |
1.91 |
1.68 |
1.35 |
0.57 |
Find the order of reaction and calculate its rate constant.
Catalytic decomposition of nitrous oxide by gold at 900°C at an initial pressure of 200 mm was 50% in 53 minutes and 73% in 100 minutes.
(i) What is the order of reaction?
(ii) How much will it decompose in 100 minutes at the same temperature but at an initial pressure of 600 mm?