The decomposition of SO2Cl2 is first order and has a rate constant of + 2.90 x 10^–4 s^–1. What is the concentration of SO2Cl2 after 840 seconds given that the initial concentration of SO2Cl2 is 0.0325M?
Well, if the decomposition of SO2Cl2 is first order, then we can use the first-order rate equation: ln([R]t/[R]0) = -kt.
Let's plug in the values we have. The initial concentration [R]0 is given as 0.0325 M, the rate constant k is 2.90 x 10^–4 s^–1, and the time t is 840 seconds.
Now, let's solve the equation:
ln([R]t/0.0325) = -(2.90 x 10^–4 s^–1)(840 s)
ln([R]t/0.0325) = -0.2436
Next, we can isolate [R]t by exponentiating both sides of the equation:
([R]t/0.0325) = e^(-0.2436)
[R]t = (0.0325) x e^(-0.2436)
[R]t ≈ 0.0325 x 0.7842
[R]t ≈ 0.0255 M
So, the concentration of SO2Cl2 after 840 seconds is approximately 0.0255 M. Just a little bit of SO2Cl2 left in the mix – it's decomposing like my hopes and dreams at a talent show.
To find the concentration of SO2Cl2 after 840 seconds, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
Plugging in the given values:
ln([A]t/0.0325) = -(2.90 x 10^–4 s^–1)(840 s)
Simplifying the equation:
ln([A]t/0.0325) = -0.2436
Next, rearrange the equation to solve for [A]t:
[A]t/0.0325 = e^(-0.2436)
Now, solve for [A]t:
[A]t = 0.0325 x e^(-0.2436)
Using a calculator, we find:
[A]t ≈ 0.0243M
Therefore, the concentration of SO2Cl2 after 840 seconds is approximately 0.0243M.
To solve this problem, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where:
[A]t is the concentration of the reactant at time t
[A]0 is the initial concentration of the reactant
k is the rate constant for the reaction
t is the time
First, we need to calculate the natural logarithm of the ratio of the concentration at time t to the initial concentration. Then we can rearrange the equation to solve for [A]t. Let's go step by step:
1. Now, let's plug in the values we know into the equation:
ln([A]t/0.0325) = -(2.90 x 10^–4 s^–1) * (840 s)
2. Rearrange the equation to solve for [A]t:
ln([A]t/0.0325) = -2.448
Using the natural logarithm properties, we can rewrite the equation as:
[A]t/0.0325 = e^(-2.448)
3. Now, solve for [A]t by multiplying both sides of the equation by 0.0325:
[A]t = 0.0325 * e^(-2.448)
4. Finally, calculate the concentration of SO2Cl2 at 840 seconds:
[A]t = 0.0325 * 0.086296
[A]t ≈ 0.0028 M
Therefore, the concentration of SO2Cl2 after 840 seconds is approximately 0.0028 M.
ln(No/N) = kt
No = 0.0325
N = ?
k = 2.90E-4
t = 840
Substitute and solve for N. Post your work if you get stuck.