Consider the data for a gas-phase reaction between A(g) and B(g). The initial ra
ID: 482679 • Letter: C
Question
Consider the data for a gas-phase reaction between A(g) and B(g). The initial rate was observed for varying partial pressures of each gas in three different trials. After determining a rate law, calculate the rate of the reaction at a partial pressure for A of PA=137 torr and a partial pressure for B of PB=55 torr. Please Explain.
Consider the data for a gas-phase reaction between Alg) and B g). The initial rate was observed for varying partial pressures of each gas in three different trials. After determining a rate law, calculate the rate of the reaction at a partial pressure for A of PA 137 torr and a partial pressure for B of PB 55 torr. Trial PA PB (torr Initial Rate (torr) (torr/sec) 100 200 50.0 400 100 50.0 300 450. 200Explanation / Answer
Given reaction is A + B ------------>
rate = k [PA]m [PB]n
From Trial I,
50= k [100]m [200]n ------------- Eq (1)
From Trial II,
50= k [100]m [400]n ------------- Eq (2)
From Trial III,
450= k [300]m [200]n ------------ Eq (3)
Eq(2)/Eq(1) gives,
50/50 = (k/k) [100/100]m [400/200]n
1 = [1]m [2]n
2o = [2]n
n= 0
---------------
Eq(3)/Eq(1) gives,
450/50 = (k/k) [300/100]m [200/200]n
9 = [3]m [1]n
32 = [3]m
m = 2
Then,
m = 2, n= 0
Therefore,
rate law is
rate = k [PA]2 [PB]0
rate = k [PA]2
Use Eq(1) to determine the value of k,
50 = k [100]m [200]n ------------- Eq (1)
50 torr/sec = k [100 torr]2 [200]0
k = 0.005 torr-1 sec-1
rate of the reaction at a partial pressure for A of PA=137 torr and a partial pressure for B of PB=55 torr
rate = k [PA]2
= (0.005 torr-1 sec-1) x (137 torr )2
= 93.8 torr/sec
Therefore,
rate of reaction = 93.8 torr/sec