Question

The decomposition of SO₂Cl₂ is first order in SO₂Cl₂ and has a rate constant of 1.38×10⁻⁴ s⁻¹ at a certain temperature.

What is the half-life for this reaction?

Answer 1

The half-life for the first-order decomposition of sulfuryl chloride (SO2Cl2) with a rate constant of 1.38×10−4 s−1 is approximately 5020 seconds or 83.67 minutes.

Step-by-step explanation:

The half-life (t1/2) of a reaction is the time it takes for half of the reactant to be consumed. For a first-order reaction, such as the decomposition of sulfuryl chloride (SO2Cl2), the half-life is determined using the equation:

t1/2 = ln(2) / k

Where k is the rate constant and ln(2) is the natural logarithm of 2 (~0.693). Given that the rate constant k for the decomposition of SO2Cl2 is 1.38×10−4 s−1, we can substitute the values into the equation:

t1/2 = ln(2) / (1.38×10−4s−1)

After performing the calculation, we find:

t1/2 = 0.693 / 1.38×10−4s−1

t1/2 = 5020.29 s. Thus, the half-life for the decomposition of SO2Cl2 is approximately 5020 seconds or 83.67 minutes.