KClO₃+6 FeSO₄+3 H₂ SO₄ → KCl+3 Fe₂(SO₄)₃+3 H₂ O
The above reaction was studied at 300 K by monitoring the concentration of FeSO₄ in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe₂(SO₄)₃ is _________ × 10⁻⁶ mol L s⁻¹ (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Get the rate from FeSO₄, then scale it by the 6 : 3 ratio in the balanced equation.

1. Rate of FeSO4 loss. The concentration falls by 10 - 8.8 = 1.2 M in half an hour, which is 1800 s. So FeSO₄ is used up at 1.2/1800 = 6.67 × 10⁻⁴ mol L⁻¹ s⁻¹.
2. Link the two rates. Divide each rate by its coefficient to get one common rate of reaction. Here FeSO₄ has coefficient 6 and Fe₂(SO₄)₃ has coefficient 3.

   -1/6d[FeSO₄]/dt = 1/3d[Fe₂(SO₄)₃]/dt

3. Put in numbers. d[Fe₂(SO₄)₃]/dt = 3/6 × 6.67 × 10⁻⁴ = 3.33 × 10⁻⁴ mol L⁻¹ s⁻¹.
4. So. The rate is 333 × 10⁻⁶ mol L⁻¹ s⁻¹, so the answer is 333.

The common slip: 667. Skipping the 3/6 factor reports the rate of FeSO₄ loss, 667 × 10⁻⁶, instead of the rate of Fe₂(SO₄)₃ formation.

The idea to keep. Rates of different species are tied together by their coefficients: divide each rate by its own coefficient and they all become equal. Also convert the time to seconds when the units ask for s⁻¹.

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