Question

The oxidation of NO (released in small amounts in the exhaust of automobiles) produces the brownish-red gas NO2 which is a component of urban air pollution. 2NO + O2 ightarrow 2NO2 The rate law for the reaction is rate = k[NO]$^2$[O2]. At 25°C, k = 7.1 x 10$^9$ L$^2$ mol$^{-2}$ s$^{-1}$. What would be the rate of the reaction if [NO] = 0.0030 mol L$^{-1}$ and [O2] = 0.027 mol L$^{-1}$?

          The oxidation of NO (released in small amounts in the exhaust of automobiles) produces the brownish-red gas NO2 which is a component of urban air pollution.
2NO + O2 
ightarrow 2NO2
The rate law for the reaction is
rate = k[NO]$^2$[O2]. At 25°C, k = 7.1 x 10$^9$ L$^2$ mol$^{-2}$ s$^{-1}$.
What would be the rate of the reaction if
[NO] = 0.0030 mol L$^{-1}$ and
[O2] = 0.027 mol L$^{-1}$?

The oxidation of NO (released in small amounts in the exhaust of automobiles) produces the brownish-red gas NO2 which is a component of urban air pollution.
2NO + O2
ightarrow 2NO2
The rate law for the reaction is
rate = k[NO]^2[O2]. At 25°C, k = 7.1 x 10^9 L^2 mol^-2 s^-1.
What would be the rate of the reaction if
[NO] = 0.0030 mol L^-1 and
[O2] = 0.027 mol L^-1?

Added by Jonathan L.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

11/22/2023

Step 1

1 * 10^9 L^2 mol^-2 s^-1 [NO] = 0.0030 mol L^-1 [O2] = 0.027 mol L^-1 Now, we can plug these values into the rate law equation: Show more…

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The oxidation of NO (released in small amounts in the exhaust of automobiles) produces the brownish-red gas NO2, which is a component of urban air pollution: 2NO + O2 -> 2NO2 The rate law for the reaction is: rate = k[NO]^2[O2] At 25°C, k = 7.1 * 10^9 L^2 mol^-2 s^-1 What would be the rate of the reaction if [NO] = 0.0030 mol L^-1 and [O2] = 0.027 mol L^-1?