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.

Chalk it up to "not focused" and work on focusing on the work at hand.