gas is entering the environment at a constant rate of 6.93 x 10^-6 g/L/day due to emission of polluting gases from thermal power plant at Kota, but at the same time it is decomposing and following fir

Question

$SO_3$ gas is entering the environment at a constant rate of 6.93 x 10^-6 g/L/day due to emission of polluting gases from thermal power plant at Kota, but at the same time it is decomposing and following first-order kinetics with half-life of 100 days. Based on these details, select the correct statement(s) from the following:

Prashant Kotian · Accepted Answer

Correct answer: A,D. At steady state: entry rate = decay rate. Entry = 6.93×10⁻⁶ g/L/day = 8.66×10⁻⁸ M/day. k = 0.693/100 = 6.93×10⁻³ day⁻¹. Steady-state C = 8.66×10⁻⁸ / 6.93×10⁻³ = 1.25×10⁻⁵ M. After 1000 days (10 half-lives), C reduces by 2⁻¹⁰ ≈ 10⁻³, giving 1.25×10⁻⁸ M.

💡 Solution & Explanation

Practice 22,000+ questions like this