Question

At the instant when NO is reacting at the rate 1.0 x10-4 mol/L.s, what is the rate at which O2 is reacting and NO2 is forming?

Answer 1

The rate at which O₂ is reacting is 5.0 x 10⁻⁵ mol/L·s, and the rate at which NO₂ is forming is 1.0 x 10⁻⁴ mol/L·s, based on the stoichiometry of the balanced chemical equation.

Step-by-step explanation:

To determine the rate at which O₂ is reacting and NO₂ is forming when NO is reacting at a rate of 1.0 x 10⁻⁴ mol/L·s, we use the stoichiometry of the balanced chemical equation:

2NO₂ (g) → 2NO(g) + O₂ (g)

According to the equation, 2 moles of NO₂ decompose to form 2 moles of NO and 1 mole of O₂ . The rates of formation of products and the rates of consumption of reactants are related by their stoichiometric coefficients. For every 2 moles of NO produced, 1 mole of O₂ is produced, so the rate of formation of NO is the same as the rate of disappearance of NO₂ , and half that rate for O₂ .

Given the rate of NO formation (1.0 x 10⁻⁴ mol/L·s), the rate of O₂ reaction is half of that, which is 5.0 x 10-5 mol/L·s. Similarly, since 2 moles of NO₂ form for every 2 moles of NO produced, the rate of NO₂ formation is also 1.0 x 10⁻⁴mol/L·s.

Answer 2

The rate at which O₂ is reacting is 5.0 x 10⁻⁵ mol/L·s, and the rate of NO₂ formation is 1.0 x 10⁻⁴ mol/L·s, following the stoichiometry of the provided reaction.

Step-by-step explanation:

The rate of a chemical reaction is typically dependent on the stoichiometry of the reaction. In the provided reaction 2NO(g) + O₂(g) → 2NO₂(g), you can see that 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂. Therefore, the rate at which NO disappears is the same rate at which NO₂ forms, and half the rate at which O₂ reacts.

Given that NO is reacting at a rate of 1.0 x 10⁻⁴ mol/L·s, O₂ would be reacting at half of this rate, which is 5.0 x 10⁻⁵ mol/L·s because it takes only 1 mole of O2 for 2 moles of NO. The formation rate of NO₂ would be the same as the disappearing rate of NO, hence 1.0 x 10⁻⁴ mol/L·s.