Chapter 12: Q12.3ACYL (page 661)

The rate law for the reaction: \({{\bf{H}}_{\bf{2}}}\) (g) + 2\({\bf{NO}}\) (g) ⟶\({\bf{N}}2{\bf{O}}\) (g) + \({{\bf{H}}_{\bf{2}}}\)O(g) has been determined to be rate = k(\({\bf{NO}}\))² (\({{\bf{H}}_{\bf{2}}}\)). What are the orders with respect to each reactant, and what is the overall order of the reaction?

Short Answer

Expert verified

The orders with respect to each reactant order in \({\bf{NO}}\)= 2; order in \({{\bf{H}}_{\bf{2}}}\)= 1; and the overall order of the reaction overall order = 3.

Step by step solution

Rate of a Reaction

The rate of reaction may be defined as the speed of the reactant reacting to obtaina product in a particular reaction at a particular time. The concentration of the reactant and product are represented in terms of mole/L.

\({\bf{rate = k}}{\left( {\bf{A}} \right)^{\bf{m}}}{\left( {\bf{B}} \right)^{\bf{n}}}\)

Explanation

Here m = 2 and n =1

\({\bf{rate = k(NO}}\left) {^{\bf{2}}} \right({\bf{H2)}}{\bf{.}}\)

Therefore:

The orders with respect to each reactant order in \({\bf{NO}}\) = 2; order in \({{\bf{H}}_{\bf{2}}}\)= 1; and the overall order of the reaction overall order = 3.

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Trial	(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)	\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls⁻¹)
1.	1.75 × 10⁻³	2.06 × 10⁻¹¹
2.	3.50 × 10⁻³	8.24 × 10⁻¹¹
3.	7.00 × 10⁻³	3.30 × 10⁻¹⁰

Trial	Time(s)	(A)
1	4.0	0.220
2	8.0	0.144
3	12.0	0.110
4	16.0	0.088
5	20.0	0.074

Time (s)	9.0	13.0	18.0	22.0	25.0
(P) (M)	1.077 × 10⁻³	1.068 × 10⁻³	1.055 × 10⁻³	1.046 × 10⁻³	1.039 × 10⁻³

Short Answer

Step by step solution

Rate of a Reaction

Explanation

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