Consider the reaction
2A (g) + B (g) --> C (g)
At 300K temperature, pressure of A and B are 2*10^5 Pa. The temperature is kept constant and let the reaction occur. After 5 mins the total pressure of the system os 3.5*10^5 Pa.
Question : When the pressure of A (g) is 1.5*10^5 Pa (during the reaction) reaction rate is 3/8rd of the initial reaction rate. Find the order of reaction relative to A and B.(A and B are whole numbers)
Method I followed :
I considered initial reaction rate as 8R and got one equation as (note that I've considered reaction order relative to A and B as x and y)
8R = (2*10^5)^x * (2*10^5)^y -->(1)
8R*3/8 = (1*10^5)^x * (1.5*10^5)^y -->(2)
Dividing 1/2
8/3 = 2^x * (1.5)^y
2*(4/3) = 2^x * (4/3)^y
How do I find x and y with the given data?
8/3 = 2^x * (4/3)^y
(4/3)^y = 8/3 * 2^-x
y ln 4/3 = ln(8/3) - x ln 2
y = (ln(8/3) - x ln2)/ln(4/3)
≈ (0.9808-0.6931x)/0.2877 = 3.4094 - 2.4094x
To find the values of x and y, we can use the information given in the question. We know that when the pressure of A is 1.5*10^5 Pa, the reaction rate is 3/8th of the initial reaction rate.
Let's assume the initial reaction rate is R. According to the question, when the pressure of A (g) is 1.5*10^5 Pa, the reaction rate is 3/8 R.
Using this information, we can set up an equation:
(1.5*10^5)^x * (2*10^5)^y = (3/8)R
Now, let's substitute the initial reaction rate in terms of R:
(1.5*10^5)^x * (2*10^5)^y = (3/8)(8R)
Simplifying, we get:
(1.5*10^5)^x * (2*10^5)^y = 3R
To find the order of the reaction relative to A and B, we need another equation. One equation is not sufficient to determine the values of both x and y.
If you have any additional information or another equation related to the reaction rate, please provide it, and we can proceed further to solve for x and y.