Rate of Reaction


The rates of consumption/disappearance, rates of production/appearance, and the rate of reaction for a chemical reaction can be related stoichiometrically for a given time interval. The relationship for a balanced, general reaction

\[ aA + bB \longrightarrow cC + dD \]

can be written such that

\[\require{color} \mathrm{rate~of~reaction} = \dfrac{1}{a} {\textcolor{red}{\dfrac{-\Delta [A]}{\Delta t}}} = \dfrac{1}{b}\textcolor{blue}{\dfrac{-\Delta [B]}{\Delta t}} = \dfrac{1}{c}\textcolor{green}{\dfrac{\Delta [C]}{\Delta t}} = \dfrac{1}{d}\textcolor{orange}{\dfrac{\Delta [D]}{\Delta t}}\]

It is important to note the following terminology definitions:

  • Rate of consumption/disappearance of \(A\) = \(\textcolor{red}{\dfrac{-\Delta [A]}{\Delta t}}\) = a number

  • Rate of consumption/disappearance of \(B\) = \(\textcolor{blue}{\dfrac{-\Delta [B]}{\Delta t}}\) = a number

  • Rate of production/formation/appearance of \(C\) = \(\textcolor{green}{\dfrac{\Delta [C]}{\Delta t}}\) = a number

  • Rate of production/formation/appearance of \(D\) = \(\textcolor{orange}{\dfrac{\Delta [D]}{\Delta t}}\) = a number

Therefore, when referring to the rate of disappearance of a reactant (e.g. A), we are referring to the decrease in the concentration of A with respect to some time interval, ΔT. (The point here is, the phrase “rate of disappearance of A” is represented by the fraction specified above). All rates are positive. Therefore, the terms involving reactants get a negative sign in front.


Stoichiometric Relationships

For the following balanced reaction

\[\mathrm{H_2} + \mathrm{I_2} \longrightarrow 2\mathrm{HI}\] The rate of consumption of H2 and I2 is half the rate of production of HI.

\[\dfrac{-\Delta [\mathrm{H_2}]}{\Delta t} = \dfrac{-\Delta [\mathrm{I_2}]}{\Delta t} = \dfrac{1}{2}\dfrac{\Delta [\mathrm{HI}]}{\Delta t}\]

This may seem strange until realizing the following:

\[\begin{align*} 2\times \dfrac{-\Delta [\mathrm{H_2}]}{\Delta t} = \dfrac{\Delta [\mathrm{HI}]}{\Delta t} \\[1ex] \end{align*}\]

By multiplying the above expression by 2, we see that the rate of appearance of HI is TWICE as fast as the rate of disappearance of H2.

Example

Suppose for the reaction above, the rate of decomposition of H2 is 0.1 mol L−1 h−1. What is the corresponding rate of reaction? What is the rate of formation of HI?

\[\begin{align*} \mathrm{rate~of~reaction} &= \dfrac{1}{1}\dfrac{-\Delta [\mathrm{H_2}]}{\Delta t} \\[1ex] &= 0.1~\mathrm{mol~L^{-1}~hr^{-1}}\\[3ex] \dfrac{1}{2}\dfrac{\Delta [\mathrm{HI}]}{\Delta t} &= \mathrm{rate~of~reaction} \\[1ex] \dfrac{\Delta [\mathrm{HI}]}{\Delta t} &= 2\left( \mathrm{rate~of~reaction} \right )\\[1ex] &= 2\left ( 0.1~\mathrm{mol~L^{-1}~hr^{-1}} \right )\\[1ex] &= 0.2~\mathrm{mol~L^{-1}~hr^{-1}} \end{align*}\]


Practice Problems


Problem 1

Write an equation that relates the rates of consumption, production, and the reaction to each other for the following chemical equation:

4NH3 (g) + 5O2 (g) ⟶ 4NO (g) + 6H2O (g)

Solution

\[\begin{align*} \mathrm{rate~of~reaction} = \dfrac{1}{4} \dfrac{-\Delta[\mathrm{NH_3}]}{\Delta t} = \dfrac{1}{5} \dfrac{-\Delta[\mathrm{O_2}]}{\Delta t} = \dfrac{1}{4} \dfrac{\Delta[\mathrm{NO}]}{\Delta t} = \dfrac{1}{6} \dfrac{\Delta[\mathrm{H_2O}]}{\Delta t} \end{align*}\]


Problem 2

In the first 10.0 s of a reaction, [I] drops from 1.000 M to 0.868 M.
What is the (a) rate of reaction in this time interval and (b) what is the rate of consumption of [H+] in this time interval? Report both answers in M s–1.

H2O2 (aq) + 3I (aq) + 2H+ (aq) ⟶ I3 (aq) + 2H2O(l)

Solution

\[\begin{align*} \mathrm{rate~of~reaction} &= \dfrac{1}{3}\dfrac{-\Delta [\mathrm{I}^-]}{\Delta t} \\[1ex] &= -\dfrac{1}{3}\dfrac{0.868~M - 1.000~M}{10.0~\mathrm{s}} = 4.40\times 10^{-3}~M~\mathrm{s^{-1}}\\[3ex] \mathrm{rate~of~H^+~consumption} &= 2(\mathrm{avg~rate~of~reaction}) \\[1ex] &= 2(4.40\times 10^{-3}~M~\mathrm{s^{-1}}) = 8.80\times 10^{-3}~M~\mathrm{s^{-1}} \end{align*}\]


Problem 3

The rate of disappearance of I2 at a particular moment during the following reaction is 1.40 × 10–2 M s–1. What is the rate of appearance (in M s–1) of HI at that moment?

H2(g) + I2(g) ⟶ 2HI(g)

Solution

\[\begin{align*} -\dfrac{1}{2}\dfrac{\Delta [\mathrm{HI}]}{\Delta t} &= -\dfrac{\Delta [\mathrm{I_2}]}{\Delta t} \\[1ex] \dfrac{\Delta [\mathrm{HI}]}{\Delta t} &= 2\left ( \dfrac{\Delta [\mathrm{I_2}]}{\Delta t} \right ) \\[1ex] &= 2\left ( 1.4\times 10^{-2}~M~\mathrm{s^{-1}} \right ) = 2.80\times 10^{-2}~M~\mathrm{s}^{-1} \end{align*}\]


Problem 4

For the following reaction, the rate of appearance of H2 is 0.345 mol L–1 min–1. Determine the following (in mol L–1 min–1)

  1. rate of reaction
  2. rate of production of N2
  3. rate of consumption of NH3

2NH3 ⟶ N2 + 3H2

Solution

Setup equation: \[\mathrm{rate~of~reaction} = \dfrac{1}{2} \dfrac{-\Delta[\mathrm{NH_3}]}{\Delta t} = \dfrac{\Delta[\mathrm{N_2}]}{\Delta t} = \dfrac{1}{3} \dfrac{\Delta[\mathrm{H_2}]}{\Delta t}\]

  1. rate of reaction \[\begin{align*} \mathrm{rate~of~reaction} &= \dfrac{1}{3} \dfrac{\Delta[\mathrm{H_2}]}{\Delta t} \\[1ex] &= \dfrac{1}{3} \left ( 0.345~\mathrm{mol~L^{-1}~min^{-1}}\right ) \\[1ex] &= 0.115 \end{align*}\]

  2. rate of production of N2 \[\begin{align*} \mathrm{rate~of~reaction} &= \dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \\[1ex] 0.115~\mathrm{mol~L^{-1}~min^{-1}} &= \end{align*}\]

  3. rate of consumption of NH3 \[\begin{align*} \mathrm{rate~of~reaction} &= \dfrac{1}{2} \dfrac{-\Delta[\mathrm{NH_3}]}{\Delta t} \\[1ex] 0.115~\mathrm{mol~L^{-1}~min^{-1}} &= \dfrac{1}{2} \dfrac{-\Delta[\mathrm{NH_3}]}{\Delta t} \\[1ex] 2\left ( 0.115~\mathrm{mol~L^{-1}~min^{-1}} \right ) &= \dfrac{-\Delta[\mathrm{NH_3}]}{\Delta t} \\[1ex] 0.230~\mathrm{mol~L^{-1}~min^{-1}} &= \end{align*}\]


Timed Assessments


Determine rates

VIDEO

Answers

Question 1

  1. 0.300 M s–1
  2. 0.450 M s–1
  3. 0.600 M s–1
  4. 0.450 M s–1

Question 2

  1. 0.119 M h–1
  2. 0.357 M h–1
  3. 0.119 M h–1
  4. 0.357 M h–1

Question 3

  1. 0.0345 mol L–1 min–1
  2. 0.104 mol L–1 min–1
  3. 0.0345 mol L–1 min–1


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