If compound A reacts with B following first order kinetics with rate constant 2.011 × 10⁻³ s⁻¹. The time taken by A (in seconds) to reduce from 7 g to 2 g will be ___________. (Nearest Integer)
[log 5=0.698, log 7=0.845, log 2=0.301]

Solution

Chemistry ma’am · 4 steps

Mass is proportional to amount of A, so plug the ratio 7/2 straight into the first order equation.

1. Mass works as concentration. For the same substance, mass is proportional to moles. So the ratio [A]₀/[A] equals the mass ratio 7/2. You do not need the molar mass.

   t = 2.303/k log[A]₀/[A] = 2.303/(2.011 × 10⁻³)log 7/2

2. Find the log. log 7/2 = log 7 - log 2 = 0.845 - 0.301 = 0.544.
3. Put in numbers. 2.303 × 0.544 = 1.2528. Then t = 1.2528/(2.011 × 10⁻³) = 622.99 s.
4. So. A takes about 623 s to fall from 7 g to 2 g. The answer is 623.

The common slip: 271. Using log 7/2 in place of ln 7/2 and dropping the 2.303 gives about 271 s, not 623.

The idea to keep. In first order kinetics only the ratio of initial to final amount matters. Masses, moles or concentrations all give the same ratio for one substance. Keep the 2.303 when you use log base 10.

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