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Consider the reaction: 2NO(g) + O₂(g) → 2NO₂(g). Suppose that at a particular moment during the reaction, nitric oxide (NO) is reacting at a rate of 0.066 m/s. At what rate is molecular oxygen (O₂) reacting?

User Pigfox
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Final answer:

The rate law for the reaction 2NO(g) + O₂(g) → 2NO₂(g) is rate = k[NO]²[O₂]. To find the rate at which O₂ is reacting, you can use the rate law equation and plug in the given rate for [NO]. The specific value for k would need to be determined from experimental data or provided in the problem.

Step-by-step explanation:

The rate law for the reaction is determined using the coefficients of the reactants in the balanced equation. In this case, the balanced equation is 2NO(g) + O₂(g) → 2NO₂(g). The rate law can be written as rate = k[NO]^x[O₂]^y, where x and y are the exponents that need to be determined. Since the coefficient of NO is 2, the exponent for [NO] is 2. The rate law would be rate = k[NO]²[O₂].

To find the rate at which O₂ is reacting, we can use the rate law equation and plug in the given rate for [NO].

[NO] rate = k[NO]²[O₂]

0.066 M/s = k(0.066 M)²[O₂]

[O₂] rate = 0.066 M/s / (k(0.066 M)²)

The specific value for k would need to be determined from experimental data or provided in the problem.

User Tri Bui
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