Final answer:
The half-life of the decomposition of PH3 at 881.00°C with a rate constant of 0.0935 s⁻¹ is approximately 7.42 seconds, corresponding to answer choice A.
Step-by-step explanation:
The student is asking about the half-life of a first-order chemical reaction, specifically the decomposition of phosphine (PH3) at a given temperature. The half-life of a first-order reaction can be determined using the formula:
t₁/₂ = ln(2) / k
Where t₁/₂ is the half-life and k is the rate constant. Substituting the given rate constant (k = 0.0935 s⁻¹) into the formula, we can calculate the half-life:
t₁/₂ = ln(2) / 0.0935 s⁻¹
t₁/₂ ≈ 7.42 s
Therefore, the correct answer is A. 7.42 s, which represents the half-life of the decomposition of PH3 at 881.00°C.