91影视

General rate law

\({\bf{Rate = k(NO}}{{\bf{)}}^m}{{\bf{(Cl}}{}_{\bf{2}}{\bf{)}}^n}\)

From the table, we can make the following equation

\(\begin{align}1.14mol{L^{ - 1}}{h^{ - 1}} &= k{(0.50)^m}{(0.50)^n}\,\,\,\,\,......(1)\\4.56mol{L^{ - 1}}{h^{ - 1}} &= k{(1.00)^m}{(0.50)^n}\,\,\,\,\,......(2)\\9.12mol{L^{ - 1}}{h^{ - 1}} &= k{(1.00)^m}{(1.00)^n}\,\,\,\,\,......(3)\end{align}\)

Cl₂remains constant, but NO is double, and the rate becomes 4 times large. so, m=2

Second-order in NO.

It is calculated by dividing equation (2) by (1)

\(\begin{align}\frac{{1.14mol{L^{ - 1}}{h^{ - 1}} = k{{(0.50)}^m}{{(0.50)}^n}}}{{4.56mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(0.50)}^n}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.25 &= {(0.5)^m}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m &= 2\end{align}\)

NO remains constant, but Cl\({}_2\) is double, and the rate becomes times large. so, \({\bf{n = 1}}\)

First-order in Cl₂

The order of Cl can be calculated by dividing equation (3) from (2)

\(\begin{align}\frac{{4.56mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(0.50)}^n}}}{{9.12mol{L^{ - 1}}{h^{ - 1}} = k{{(1.00)}^m}{{(1.00)}^n}}}\\0.5 &= {(0.5)^m}\\m &= 1\end{align}\)

The rate constant is the proportionality constant in the equation that expresses the relationship between the rate of a chemical reaction and the concentration of reacting substances.

It can be calculated as

\(\begin{align}k &= \frac{{rate}}{{{{(NO)}^m}{{(Cl{}_2)}^n}}}\\ &= \frac{{1.14mol\,{L^{ - 1}}\,{h^{ - 1}}}}{{{{(0.50\,mol\,{L^{ - 1}})}^2}(0.50\,mol\,{L^{ - 1}})}}\\ &= 9.1\,{L^2}\,mo{l^{ - 2\,}}{h^{ - 1}}\end{align}\)