The decomposition of N2O5 proceeds according to the following equation
2 N2O5(g) → 4 NO2(g) + O2(g).
The rate constant, k, for the reaction at 298 K is 2.20 x 10-3 min-1. If a reaction is set up in a vessel at 298 K, with an initial concentration of 0.343 mol L-1; what is the concentration of reactant N2O5(g) in mol/L after 42 minutes?
The units of min^-1 tell you that the reaction is first order.
So ln(No/N) = kt.
No is starting.
N is end
k from problem
t = 42 min
298/0.343 = 868
868/42 = 21.000
Nope. You didn't substitute. And you know the answer can't be right. You start with 0.343 mol, it decomposes and you have more than you started with after 43 mn. I don't think so.
To find the concentration of reactant N2O5(g) after 42 minutes, we can use the integrated rate law for a first-order reaction:
ln[A] = -kt + ln[A]₀
where [A] is the concentration of reactant at time t, k is the rate constant, t is time, and [A]₀ is the initial concentration of the reactant.
Given the rate constant (k = 2.20 x 10^(-3) min^(-1)), the initial concentration ([A]₀ = 0.343 mol L^(-1)), and the time (t = 42 minutes), we can substitute these values into the equation to find the concentration of N2O5(g) after 42 minutes.
ln[A] = -kt + ln[A]₀
ln[A] = -(2.20 x 10^(-3) min^(-1))(42 min) + ln(0.343 mol L^(-1))
Using a calculator, we first calculate the value of -kt:
-(2.20 x 10^(-3) min^(-1))(42 min) = -0.0924
Next, we find the natural logarithm of the initial concentration:
ln(0.343 mol L^(-1)) ≈ -1.072
Substituting these values into the equation:
ln[A] ≈ -0.0924 + (-1.072)
ln[A] ≈ -1.164
To find the concentration of N2O5(g), we need to convert back from the logarithmic form to the regular concentration form:
[A] = e^ln[A]
Using a calculator:
[A] ≈ e^(-1.164)
[A] ≈ 0.312 mol L^(-1)
Therefore, the concentration of N2O5(g) after 42 minutes is approximately 0.312 mol L^(-1).