Question

The compound SO2Cl2 decomposes in a first-order reaction SO2Cl2(g) = SO2(g)+Cl2(g) that has a half-life of 1.47*10⁴ s at 600. K. If you begin with 1.3*10^-3 mol of pure SO2Cl2 in a 4.0-L flask, calculate at what time the amount of SO2Cl2 will be 1.6*10⁻⁴ mol.

Answer 1

The amount of SO2Cl2 will be 1.6*10⁻⁴ mol after 1.47*10⁴ seconds.

Step-by-step explanation:

To find the time at which the amount of SO2Cl2 will be 1.6*10⁻⁴ mol, we can use the first-order rate equation:

ln([SO2Cl2]t/[SO2Cl2]0) = -kt

We can rearrange this equation to solve for t:

t = (-1/k) * ln([SO2Cl2]t/[SO2Cl2]0)

Given that the half-life is 1.47*10⁴ s and the initial concentration is 1.3*10^-3 mol, we can substitute these values into the equation:

t = (-1/0.693) * ln(1.6*10⁻⁴/1.3*10^-3) = 1.47*10⁴ s

Therefore, the amount of SO2Cl2 will be 1.6*10⁻⁴ mol after 1.47*10⁴ seconds.