(a) The order with respect to A is 4, the order with respect to B is 3.
(b) The rate law is rate = k[A]^4[B]^3.
(c) k = (6.25 x 10^-3)/(0.0500^4 x 0.0500^3) = 0.5 x 10^6.
For the reaction 4A (g)+3B (g)=C(g) the following data were obtained at constant temperature:
Experiment-
1. Initial (A) (mole/L)
0.0500, 01000, 01000 , 0.0500
2. Initial (B) (mole/L)
0.0500, 0.0500, 0.1000, 0.0500
3. Initial (C)(mole/L)
0.0100, 0.0100, 0.0100, 0.0200
Initial rate (mole/L)
6.25×10^-3, 1.25×10^-2, 5.00×10^-2, 6.25×10^-3
(a) what is the order with respect to each reactant?
(b) write the rate law?
(c) calculate k (using the data from experiment 1)
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