At the instant oxygen is reacting at the rate 5.0 x 10-4 mol/L*s, at what rate is NO reacting and NO2 forming?
this is a method of initial rate problem. i don't know what to do
the chemical equation is
2 NO(g) + O2 (g) ---> 2 NO2 (g)
Both of them have a 2 as a coefficent, so wouldn't the rate be twice that of O2?
i see what your saying. so multiply the rate by two to get the rate of NO? but what is NO2 forming
Won't the rate of disappearance of NO be the same as the rate of formation of NO2 since both of the coefficients are the same?
To solve this method of initial rate problem, you can use the stoichiometry of the balanced chemical equation. Here's how you can approach it step by step:
1. Identify the balanced chemical equation: 2 NO(g) + O2(g) → 2 NO2(g)
2. Determine the stoichiometric relationship between the reactants and products: According to the balanced equation, each mole of O2 reacts with 2 moles of NO to produce 2 moles of NO2.
3. Use the stoichiometric relationship to relate the rate of the given reaction (oxygen) to the unknown rate of the other reaction (NO and NO2 formation).
4. The rate expression for the given reaction is:
rate = -1/2 * Δ[O2]/Δt = 5.0 x 10^-4 mol/(L*s), where Δ[O2]/Δt is the change in concentration of O2 per unit time.
5. Based on the stoichiometry, the rate of NO formation would be half the rate of oxygen consumption:
rate(NO formation) = 1/2 * rate(O2) = 1/2 * 5.0 x 10^-4 mol/(L*s).
6. Similarly, the rate of NO2 formation is also half the rate of oxygen consumption:
rate(NO2 formation) = 1/2 * rate(O2) = 1/2 * 5.0 x 10^-4 mol/(L*s).
Therefore, the rate of NO formation and NO2 formation can be calculated as 2.5 x 10^-4 mol/(L*s).