Question
Question: In the following table, at different concentrations of A and B the values of rate, R are given \...
In the following table, at different concentrations of A and B the values of rate, R are given
[A], mol L−1 | [B], mol L−1 | R, mol L−1 s−1 |
---|---|---|
0.1 | 0.2 | 1×10−3 |
0.2 | 0.4 | 8×10−3 |
A. 1
B. 2
C. 3
D. 221
Solution
For this we have to use the rate law equation. We will get two equations on substituting the given data. We can compare the two equations to find the order with respect to each reactant. The overall order is calculated by adding the order with respect to A and B.
Formula used: R=k[A]a[B]b here R is rate, k is rate constant and ‘a’ and ‘b’ are order with respect to A and B respectively.
Complete step by step answer:
Let the order with respect to reactant A is ‘a’ and the order with respect to reactant B is ‘b’.
Let us put the first set of data in the formula. We will get,
1×10−3=k[0.1]a[0.2]b
Now for the second set of data we will get,
8×10−3=k[0.2]a[0.4]b
Now we will divide both the equation and we will get,
8×10−31×10−3=k[0.2]a[0.4]bk[0.1]a[0.2]b
We can combine similar power and write the equation after simplification as:
81=[21]a×[21]b
We know that when base is same then power can be added, so we will add ‘a’ and ‘b’ over the common base as:
81=[21]a+b
We can also write 8 as the cube of 2, we will get the equation:
[21]3=[21]a+b
Since the base on both sides is the same, we can compare the powers so, a+ b=3. Sum of orders with respect to A and B gives us the overall order.
Hence the overall order is 3 and the correct option is C.
Note:
‘k’ is known as the rate constant. In the question we have kept the rate constant the same for both sets of data because a rate constant depends upon the nature of reaction and temperature. If both of them are constant then k will also be constant.