In the presence of a tungsten catalyst at high temperature, the decomposition of ammonia to nitrogen and hydrogen is a zero order process. If the rate constant at a particular temperature us 3.7x10^-6 mol. How long will it take for an ammonia concentration to drop from an initial concentration of 5 .0 x10^-4 M to 5.0x10^-5 M? What is the half life of the reaction under these conditions?
Write and balance the equation.
[A] = [Ao]-akt
[Ao] is the beginning cncn.
k is given
t is time.
a is the coefficient in the balanced equation.
t1/2 = [Ao]/2ak
To solve this problem, we need to use the rate equation for a zero-order reaction and integrate it to find an expression for the reaction time and half-life.
In a zero-order reaction, the rate is independent of the concentration of the reactant. The rate equation for a zero-order reaction can be written as:
rate = k
Where:
rate is the rate of the reaction (concentration change per unit time)
k is the rate constant
To find how long it will take for the ammonia concentration to drop from an initial concentration of 5.0 x 10^-4 M to 5.0 x 10^-5 M, we need to calculate the time it takes for this concentration change to occur.
First, let's calculate the concentration change:
Concentration change = Initial concentration - Final concentration
= (5.0 x 10^-4 M) - (5.0 x 10^-5 M)
= 4.5 x 10^-4 M
Now, we can use the rate equation to find the time it takes for this concentration change to occur. Rearranging the equation, we have:
rate = k
k = rate
t = concentration change / rate
= (4.5 x 10^-4 M) / (3.7 x 10^-6 mol)
Calculating this value, we get:
t = (4.5 x 10^-4 M) / (3.7 x 10^-6 mol)
= 121.6 s
So, it will take approximately 121.6 seconds for the ammonia concentration to drop from 5.0 x 10^-4 M to 5.0 x 10^-5 M.
Now, let's calculate the half-life of the reaction. The half-life is the time it takes for the concentration to reduce to half its initial value.
For a zero-order reaction, the half-life can be calculated using the expression:
t(1/2) = initial concentration / (2 * rate)
Substituting the values, we have:
t(1/2) = (5.0 x 10^-4 M) / (2 * 3.7 x 10^-6 mol)
Calculating this value, we get:
t(1/2) = (5.0 x 10^-4 M) / (7.4 x 10^-6 mol)
= 67.6 s
So, the half-life of the reaction under these conditions is approximately 67.6 seconds.