Final answer:
b. .093. By multiplying the rate of water appearance (0.14 M/s) by the ratio of 2/3, we find that NO₂ appears at a rate of 0.093 M/s.
Step-by-step explanation:
The question is asking to determine the rate at which NO₂ appears based on the rate at which water (H₂O) is formed in the given chemical reaction. The balanced chemical equation provided is 4NH₃ + 7O₂ → 4NO₂ + 6H₂O. If water is appearing at a rate of 0.14 M/s, we must consider the stoichiometric ratio between H₂O and NO₂, which is 6:4 or 3:2. Since each 3 moles of water produced are associated with 2 moles of NO₂ being produced, we must find the equivalent rate for NO₂. We calculate this by multiplying the given rate of water appearance, 0.14 M/s, by the ratio of the stoichiometric coefficients (2/3). The rate at which NO₂ appears is calculated based on the stoichiometric ratio between H₂O and NO₂ from the balanced equation.
To get the rate at which NO₂ appears, we perform the calculation: (0.14 M/s) × (2/3) = 0.093 M/s. Therefore, the rate at which NO₂ appears is 0.093 M/s, which corresponds to option b. 0.093.