The following reaction is first order, C2H6==> 2CH3. If the rate constant is equal to 5.5 x 10-4 s^-1 at 1000 K, how long will it take for 0.35 mol of C2H6 in a 1.00 L container to decrease to 0.20 mol in the same container?
Thank you very much
I'm not surprised. Why did you substitute 1000 for k? The problem states k = 5.5 x 10^-4 s^-1. You substituted 1000 Kelvin which is the temperature.
ok i got i now thank you very much
ln(No/N)
=ln(0.35/0.20)
= 0.559
0.559/5.5*10-4
=1017.48
=17mins
ln(No/N) = kt
ln(0.35/0.20) = kt
Substitute for k and calculate t. The calculated time will be in seconds.
ln(No/N) = kt
ln(0.35/0.20) = kt
ln(0.35/0.20)=1000*t
ln(0.35/0.20)/(1000)=5.59
but that answer is not on my worksheet im i doing something wrong?
oh haha i didn't notice i feel dumb now
i am trying to find tis answer but I m getting different result
To solve this problem, we can use the first-order rate equation:
rate = k * [C2H6]
Where:
rate = rate of reaction
k = rate constant
[C2H6] = concentration of C2H6
Given:
Rate constant, k = 5.5 x 10^-4 s^-1
Initial concentration, [C2H6]₀ = 0.35 mol/L
Final concentration, [C2H6] = 0.20 mol/L
We need to find the time it takes for the concentration of C2H6 to decrease from 0.35 mol/L to 0.20 mol/L.
First, let's calculate the rate of the reaction at the initial and final concentrations:
Rate at initial concentration, rate₀ = k * [C2H6]₀
Rate at final concentration, rate = k * [C2H6]
We know that the rate is inversely proportional to the concentration. Therefore, we can set up the following ratio:
rate / rate₀ = [C2H6] / [C2H6]₀
Substituting the values:
k * [C2H6] / (k * [C2H6]₀) = 0.20 mol/L / 0.35 mol/L
Simplifying:
[C2H6] / [C2H6]₀ = 0.20 mol/L / 0.35 mol/L
Now we can solve for [C2H6]:
[C2H6] = 0.20 mol/L / 0.35 mol/L * [C2H6]₀
[C2H6] = 0.20/0.35 * 0.35 mol/L
[C2H6] = 0.20 mol/L
Now, we need to find the time it takes for the concentration of C2H6 to decrease to 0.20 mol/L.
Using the first-order rate equation:
rate = k * [C2H6]
5.5 x 10^-4 s^-1 = k * 0.20 mol/L
Now, we solve for the time, t:
t = ln([C2H6]₀ / [C2H6]) / k
t = ln(0.35 mol/L / 0.20 mol/L) / (5.5 x 10^-4 s^-1)
Calculating the time:
t = ln(1.75) / (5.5 x 10^-4 s^-1)
t ≈ 9.86 s
Therefore, it will take approximately 9.86 seconds for the concentration of C2H6 to decrease from 0.35 mol/L to 0.20 mol/L.