The reaction 2 NO(g) +O₂(g) $\rightarrow$ 2 NO₂(g) has the following rate law: Rate =k[O₂][NO]².If the concentration of NO is reduced by a factor of two while the concentration of O₂ is doubled, the rate will

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The rate law is given as Rate = k[O2][NO]^2. If [NO] is halved, its effect on the rate is (1/2)^2 = 1/4. If [O2] is doubled, its effect on the rate is 2. Multiplying these effects together, 1/4 * 2 = 1/2, so the rate decreases by a factor of one-half.

The Reaction 2 NO(g) +O2(g) →\rightarrow→ 2 NO2(g) Has the Following Rate Law: Rate =K[O2][NO]2

The Reaction 2 NO(g) +O₂(g) $\rightarrow$ 2 NO₂(g) Has the Following Rate Law: Rate =K[O₂][NO]²