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Question
Compounds ‘A’ and ‘B’ react according to the following chemical equation.
\[\ce{A(g) + 2B(g) -> 2C(g)}\]
Concentration of either ‘A’ or ‘B’ were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate equations for this reaction.
| Experiment | Initial concentration of [A]/mol L–¹ |
Initial concentration of [B]/mol L–¹ |
Initial rate of formation of [C]/mol L–¹ s–¹ |
| 1. | 0.30 | 0.30 | 0.10 |
| 2. | 0.30 | 0.60 | 0.40 |
| 3. | 0.60 | 0.30 | 0.20 |
Options
Rate = k[A]2[B]
Rate = k[A][B]2
Rate = k[A][B]
Rate = k[A]2[B]0
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Solution 1
Rate = k[A][B]2
Explanation:
Rate = k[A]x[B]y
When concentration of B is doubled keeping the concentration of A constant, the rate of formation of C increases by a factor of four. This indicates that the rate of reactions depends upon the square of concentration of B. When concentration of A is doubled, the rate of formation of C also doubles from the initial value. This shows that the rate depends on first power of concentration of A. Hence Rate= k[A]x[B]y.
Solution 2
Let order of A and B be x and y.
r = k [A]x [B]y
0.1 = k (0.3)x (0.3)y ...(1)
0.4 = 0.1 = k (0.3)x (0.6)y ...(2)
0.2 = 0.1 = k (0.6)x (0.3)y ...(3)
Divide 2 by 1
`0.4/0.1 = (0.6)^"y"/(0.3)^"y"`
Divide 3 by 1
`0.2/0.1 = (0.6)^"x"/(0.3)^"x"`
Hence Rate law is
r = k [A]1 [B]2
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