What is the partial pressure of radon if the total pressure is 780 torr and the
1 answer:
Answer:
0.03atm
Explanation:
Given parameters:
Total pressure = 780torr
Partial pressure of water vapor = 1.0atm
Unknown:
Partial pressure of radon = ?
Solution:
A sound knowledge of Dalton's law of partial pressure will help solve this problem.
The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".
Mathematically;
P = P
+ P
+ P
Since the total pressure is 780torr, convert this to atm;
760torr = 1 atm
780torr = atm = 1.03atm
For this problem;
Total pressure = Partial pressure of radon + Partial pressure of water vapor
1.03 = Partial pressure of radon + 1.0
Partial pressure of radon = 1.03 - 1.00 = 0.03atm
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Answer:
(a) The rate of disappearance of is:
M/s
(b) The value of rate constant is: 0.83036
(c) The units of rate constant is:
(d) The rate will increase by a factor of 3.24
Explanation:
The rate of a reaction can be expressed in terms of the concentrations of the reactants and products in accordance with the balanced equation.
For the given reaction:
rate = =
=
-----(1)
According to the question, the reaction is second order in NO and first order in .
Then we can say that, rate = k -----(2)
where k is the rate constant.
The rate of disappearance of NO is given:
=
M/s.
(a) From (1), we can get the rate of disappearance of .
Rate of disappearance of =
= (0.5)*(
) M/s =
M/s.
(b) The rate of the reaction can be obtained from (1).
rate = = (0.5)*(
)
rate = M/s
The value of rate constant can be obtained by using (2).
rate constant = k =
k = = 0.83036
(c) The units of the rate constant can be obtained from (2).
k =
Substituting the units of rate as M/s and concentrations as M, we get:
=
(d) The reaction is second order in NO. Rate is proportional to square of the concentration of NO.
If the concentration of NO increases by a factor of 1.8, the rate will increase by a factor of = 3.24