Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g) + O2(g) → 2 SO3(g) Given the following balanced equation, if the rate of O2 loss is 3.64 × 10-3 M/s, what is the rate of formation of SO3? 2 SO2(g) + O2(g) → 2 SO3(g) 3.64 × 10-3 M/s 6.60 × 10-2 M/s 3.99 × 10-3 M/s 7.28 × 10-3 M/s 1.82 × 10-3 M/s

Rate of formation of SO₃ = 7.28 x 10⁻³ M/s

Explanation:

According to equation   2 SO₂(g) + O₂(g) → 2 SO₃(g)

Rate of disappearance of reactants = rate of appearance of products

                     ⇒  -----------------------------(1)

    Given that the rate of disappearance of oxygen = = 3.64 x 10⁻³ M/s

             So the rate of formation of SO₃ = ?

from equation (1) we can write

                                ⇒ = 2 x 3.64 x 10⁻³ M/s

                                ⇒ = 7.28 x 10⁻³ M/s

∴ So the rate of formation of SO₃ = 7.28 x 10⁻³ M/s

there is none because earth is stationary and flat.