NO₂ required for a reaction is produced by decomposition of N₂ O₅ in CCl₄ as by equation
2 N₂ O_(5( g)) → 4 NO_(2( g))+O_(2( g))
The initial concentration of N₂ O₅ is 3 mol L⁻¹ and it is 2.75 mol L⁻¹ after 30 minutes.
The rate of formation of NO₂ is x × 10⁻³ mol L⁻¹ min⁻¹, value of x is _________. (nearest integer)

Solution

Chemistry ma’am · 4 steps

Find how fast N₂O₅ disappears, then scale by the 4 : 2 ratio to get NO₂.

1. Rate of loss. N₂O₅ falls from 3 to 2.75 mol L⁻¹, a drop of 0.25 mol L⁻¹ in 30 minutes. So it disappears at 0.25/30 = 8.33 × 10⁻³ mol L⁻¹ min⁻¹.
2. Link the species. Divide each rate by that species' coefficient in the equation and set them equal. For every 2 mol of N₂O₅ used up, 4 mol of NO₂ form.

   -1/2Δ[N₂O₅]/(Δ t) = 1/4Δ[NO₂]/(Δ t)

3. Rate of NO2. Δ[NO₂]/(Δ t) = 4/2 × 8.33 × 10⁻³ = 16.67 × 10⁻³ mol L⁻¹ min⁻¹.
4. So. x = 16.67, which rounds to 17.

The common slip: 8. Reporting the rate at which N₂O₅ disappears and leaving out the factor of 4/2 gives 8, not 17.

The idea to keep. Divide each rate by its coefficient in the balanced equation and they all become equal. Use that to convert the rate of one species into the rate of any other.

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