Consider the following reaction:
2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g)
(b) If the rate constant for this reaction at 1000 K is 6.0 × 10⁴ M^-2 s^-1, what is the reaction rate when [NO] = 0.035 M and [H₂] = 0.015 M?
(c) What is the reaction rate at 1000 K when the concentration of NO is increased to 0.10 M, while the concentration of H₂ is 0.010 M?

Verified step by step guidance

1. Identify the rate law for the given reaction. Since the reaction is 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g), the rate law can be expressed as: rate = k[NO]^m[H2]^n, where k is the rate constant, and m and n are the orders of the reaction with respect to NO and H2, respectively.
2. Determine the reaction orders (m and n) from the stoichiometry of the balanced equation. For this reaction, the stoichiometry suggests that m = 2 and n = 2, assuming the reaction is elementary.
3. Use the rate law to calculate the reaction rate for part (b). Substitute the given values into the rate law: rate = (6.0 × 10^4 M^-2 s^-1) × (0.035 M)^2 × (0.015 M)^2.
4. Calculate the reaction rate for part (c) using the same rate law. Substitute the new concentration values into the rate law: rate = (6.0 × 10^4 M^-2 s^-1) × (0.10 M)^2 × (0.010 M)^2.
5. Compare the reaction rates from parts (b) and (c) to understand the effect of changing concentrations on the rate of the reaction. Consider how the increase in [NO] and decrease in [H2] affect the overall rate.