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Question
For the reaction: \[\ce{2H2 + 2NO ⇌ 2H2O + N2}\] , the following rate data was obtained:
| S.No | [NO] mol L-1 | [H2] mol L-1 | Rate: mol L-1sec-1 |
| 1 | 0.40 | 0.40 | 4.6 × 10-3 |
| 2 | 0.80 | 0.40 | 18.4 × 10-3 |
| 3 | 0.40 | 0.80 | 9.2 × 10-3 |
Calculate the following:
(1) The overall order of a reaction.
(2) The rate law.
(3) The value of rate constant (k).
Sum
Solution
Let rate law is
r = k[H2]x [NO]y
4.6 × 10-3 = k (0.4)x (0.4)y ....(1)
18.4 × 10-3 = k (0.4)x (0.80)y ....(2)
9.2 × 10-3 = k (0.8)x (0.4)y .....(3)
Dividing equation (1) by equation (2),we have
`1/4 = (1/2)^"y"`
y = 2
Dividing equation (1) by equation (3), we have
`1/2 = (1/2)^"x"`
x = 1
Therefore,
Order w.r.t.H2 = 1
Order w.e.t. NO = 2
Overall order = 1 + 2 = 3
Rate law r = k[H2]1 [NO]2
From equation (1)
4.6 × 10-3 = k (0.4)1 (0.4)2
4.6 × 10-3 = k × 4 × 4 × 4 × 10-3
k = `4.6/64`
k = 0.07 mol-2 L2 sec-1
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Order of a Reaction - Order as an Experimental Quantity
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