The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___________ × 10⁻³ min⁻¹. (Nearest integer)
[Use : ln 10 = 2.303; log₁₀ 3 = 0.477; property
of logarithm : log xʸ = y log x]

Solution

Chemistry ma’am · 4 steps

Rate proportional to amount means first order, so use the log form with 90% left after one minute.

1. Spot the order. The rate is proportional to the amount of virus, so this is a first-order process. If 10% is inactivated, 90% is still left after t = 1 min.

   k = 2.303/t log[A]₀/[A] = 2.303/1 log 100/90

2. Work out the log. log 10/9 = log 10 - log 3² = 1 - 2 × 0.477 = 0.046.
3. Put in numbers. k = 2.303 × 0.046 = 0.1059 min⁻¹ = 105.9 × 10⁻³ min⁻¹.
4. So. The rate constant is about 106 × 10⁻³ min⁻¹, so the answer is 106.

The common slip: 46. Leaving out the factor 2.303 and using log 10/9 = 0.046 directly as k gives 46, not 106.

The idea to keep. In a first-order formula, always put in the amount that is still left, not the amount used up. Then ln = 2.303 log turns it into a log you can find from the values given.

Answer from the NTA final answer key. Solution by PrepWiser.