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)
(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.
To determine the order with respect to each reactant, we can examine how the initial rate changes when the concentration of each reactant is changed while keeping the concentration of the other reactants constant.
(a) Order with respect to A:
From experiments 1 and 3, we can see that when the concentration of A is doubled (from 0.0500 to 0.0100), the initial rate stays the same (6.25×10^-3). Therefore, the order with respect to A is zero.
Order with respect to B:
From experiments 2 and 4, we can see that when the concentration of B is doubled (from 0.0500 to 0.1000), the initial rate is quadrupled (from 1.25×10^-2 to 5.00×10^-2). This indicates that the rate is directly proportional to the concentration of B, so the order with respect to B is one.
Order with respect to C:
From experiments 3 and 4, we can see that when the concentration of C is doubled (from 0.0100 to 0.0200), the initial rate stays the same (6.25×10^-3). Therefore, the order with respect to C is zero.
(b) Rate law:
Based on the order determined above, the rate law can be written as:
Rate = k [A]^0 [B]^1 [C]^0
Since the order with respect to both A and C is zero, their concentrations do not affect the rate. The overall rate is solely determined by the concentration of B.
(c) Calculating k:
Using the data from experiment 1 (where the concentration of B is constant), we can calculate k by rearranging the rate law equation:
Rate = k [A]^0 [B]^1 [C]^0
k = Rate / [B]
From experiment 1, the initial rate is 6.25 × 10^-3 mole/L, and the concentration of B is 0.0500 mole/L. Therefore, we can calculate k as follows:
k = (6.25 × 10^-3) / (0.0500)
k = 0.125 mole^(-1) L
The value of k is 0.125 mole^(-1) L.
To determine the order of a reaction with respect to each reactant, we need to use the experimental data. The order of a reactant is the exponent to which its concentration is raised in the rate equation.
(a) To find the order with respect to each reactant, we will compare the initial rates from different experiments while keeping the concentration of one reactant constant and varying the concentration of another reactant.
For the reaction 4A (g) + 3B (g) = C (g), let's compare Experiments 1 and 2 while keeping the concentration of B (mole/L) constant at 0.0500.
Experiment 1:
Initial rate = 6.25×10^-3 (mole/L)
Experiment 2:
Initial rate = 1.25×10^-2 (mole/L)
We can see that as the concentration of A (mole/L) doubles (from 0.0500 to 0.1000), the initial rate also doubles. This indicates that the reaction is first-order with respect to A.
Now, let's compare Experiments 1 and 3 while keeping the concentration of A (mole/L) constant at 0.0500.
Experiment 1:
Initial rate = 6.25×10^-3 (mole/L)
Experiment 3:
Initial rate = 5.00×10^-2 (mole/L)
We can see that as the concentration of B (mole/L) doubles (from 0.0500 to 0.1000), the initial rate increases by a factor of 8 (from 6.25×10^-3 to 5.00×10^-2). This indicates that the reaction is second-order with respect to B.
Therefore, the order with respect to A is 1 and the order with respect to B is 2.
(b) The rate law can be expressed using the determined orders:
Rate = k [A]^1 [B]^2
(c) To calculate the rate constant (k) using the data from Experiment 1, we will substitute the values from Experiment 1 into the rate equation:
Rate (Exp. 1) = k [A (Exp. 1)]^1 [B (Exp. 1)]^2
6.25×10^-3 = k (0.0500)^1 (0.0500)^2
Simplifying:
6.25×10^-3 = k (0.0500)(0.0025)
Dividing both sides by (0.0500)(0.0025):
k = 6.25×10^-3 / (0.0500)(0.0025)
k ≈ 50
Therefore, the rate constant (k) is approximately 50 (mole/L)^-2 sec^-1.