Q30 E The annual production of \({\bf{... [FREE SOLUTION]

Chapter 12: Q30 E (page 705)

The annual production of \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\) in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.
\(\begin{align}\left( a \right){\bf{ }}4N{H_3}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}5{O_2}{\bf{ }}(g) \to 4NO\left( g \right){\bf{ }} + {\bf{ }}6{H_2}O\left( g \right)\\\left( b \right){\bf{ }}2NO\left( g \right){\bf{ }} + {\bf{ }}{O_{2{\bf{ }}}}(g) \to 2N{O_{2{\bf{ }}}}\left( g \right)\\\left( c \right){\bf{ }}3N{O_2}{\bf{ }}\left( g \right){\bf{ }} + {\bf{ }}{H_2}O(l) \to 2HN{O_3}(aq) + NO(g)\end{align}\)
The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in \({{\bf{O}}_{\bf{2}}}\), what is the rate of formation of \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is \({\bf{5}}{\bf{.8 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{ L}}{{\bf{ }}^{\bf{2}}}{\bf{ mo}}{{\bf{l}}^{{\bf{ - 2}}}}{\bf{ s}}{{\bf{ }}^{{\bf{ - 1}}}}\).

Short Answer

Expert verified

Rate of formation of \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) is \({\bf{1}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{mi}}{{\bf{n}}^{{\bf{ - 1}}}}\)

Step by step solution

Rate law for formation of \({\bf{HN}}{{\bf{O}}_{\bf{3}}}\)

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

Therefore, the rate law for equation (b)

Rate \({\bf{ = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(O}}{}_{\bf{2}}{\bf{)}}\)

Where k is rate constant.

The rate constant is the proportionality constant in rate equation.

Rate of reaction

Rate of formation of NO₂ can be calculated as

\({\bf{Rate of reaction = k(NO}}{{\bf{)}}^{\bf{2}}}{\bf{(}}{{\bf{O}}_{\bf{2}}}{\bf{)}}\)

\(\begin{align}rate\;of\;reaction &= 5.8 \times {10^{ - 6}}{L^2}mo{l^2}{s^{ - 1}}{(0.75mol{L^{ - 1}})^2}(0.50mol/L)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; &= 1.6 \times {10^{ - 6}}mol{L^{ - 1}}{\min ^{ - 1}}\end{align}\)

Thus, the rate of formation of \({\bf{N}}{{\bf{O}}_{\bf{2}}}\) is \({\bf{1}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 6}}}}{\bf{mol}}{{\bf{L}}^{{\bf{ - 1}}}}{\bf{mi}}{{\bf{n}}^{{\bf{ - 1}}}}\)