Chemistry · Ch 7 — Chemical Kinetics

Stoichiometry and Rate of a Reaction

7.1.1

Stoichiometry and Rate of a Reaction

Why stoichiometry complicates a single 'the' rate. For $A\rightarrow B$ (matching coefficients on both sides), the rate of A's disappearance and B's appearance are numerically identical, so there is no ambiguity in calling either one 'the rate'. But consider $A\rightarrow2B$ : for every mole of A consumed, TWO moles of B appear, so B's concentration climbs twice as fast as A's falls:

\text{Rate}=\frac{+d[B]}{dt}=2\left(\frac{-d[A]}{dt}\right)\quad\text{equivalently}\quad\text{Rate}=\frac{-d[A]}{dt}=\frac{1}{2}\frac{d[B]}{dt}

If you tracked only $-d[A]/dt$ and called THAT 'the rate', you would get a different number than if you tracked $d[B]/dt$ and called that the rate -- the two numbers differ by exactly the factor 2. This is clearly unsatisfactory; chemists need a SINGLE unambiguous number for 'the rate of the reaction', independent of which species happened to be measured.

The general fix. For any balanced reaction $xA+yB\rightarrow lC+mD$ , dividing each species' own rate of concentration change by ITS OWN stoichiometric coefficient produces a single common value:

$\text{Rate}=-\frac{1}{x}\frac{d[A]}{dt}=-\frac{1}{y}\frac{d[B]}{dt}=\frac{1}{l}\frac{d[C]}{dt}=\frac{1}{m}\frac{d[D]}{dt}$ …

Figure 7.1Change in concentration of A and B for the reaction A -> B

What this figure shows. A schematic (not-to-scale) graph of concentration (M, y-axis, 0 to 1.0) against time in minutes (x-axis, 0 to 120). Two curves are drawn: a pink curve for [A] starting at 1.0 M and falling smoothly toward 0 as time increases, and a green curve for [B] starting at 0 and rising smoothly toward 1.0 M, mirroring the fall of [A]. Along the top of the plot, a row of seven conical flasks is drawn, each containing a mix of pink and green dots in a ratio that shifts left-to-right from mostly-pink (early reaction, mostly unreacted A) to mostly-green (late reaction, mostly converted to B), visually tying the curve to the …

Misc example-1Example 1 -- rate expressions for the oxidation of NO

Worked out. For $2NO(g)+O_2(g) \rightarrow 2NO_2(g)$ : (a) express the rate in terms of the changes in concentration of NO, $O_2$ and $NO_2$ ; (b) at a particular instant, $[O_2]$ is decreasing at $0.2\ \text{mol L}^{-1}\text{s}^{-1}$ -- at what rate is $[NO_2]$ increasing at that instant? Book's solution: (a) $\text{Rate}=-\dfrac{1}{2}\dfrac{d[NO]}{dt}=-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}$ . (b) From $-\dfrac{d[O_2]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}$ , $\dfrac{d[NO_2]}{dt}=2\times\left(-\dfrac{d[O_2]}{dt}\right)=2\times0.2=0.4\ \t …

Misc evaluate-yourself-1Evaluate Yourself 1 -- rate expressions and N2O5 decomposition

Worked out. Book's practice box (no printed solution). (1) Write the rate expression for (i) $3A+5B_2 \rightarrow 4CD_2$ and (ii) $X_2+Y_2 \rightarrow 2XY$ , assuming both are elementary reactions. (2) $N_2O_5(g)$ decomposes to $NO_2(g)$ and $O_2(g)$ ; at a particular instant $N_2O_5$ disappears at $2.5\times10^{-3}\ \text{mol dm}^{-3}\text{s}^{-1}$ . At what rates are $NO_2$ and $O_2$ formed, and what is the rate of the reaction? Working it through: (1i) $\text{Rate}=-\dfrac{1}{3}\dfrac{d[A]}{dt}=-\dfrac{1}{5}\dfrac{d[B_2]}{dt}=\dfrac{1}{4}\dfrac{d[CD_2]}{dt}$ ; (1ii) $\text{Rate}=-\dfrac{d[X_2]}{dt}=-\dfrac{d[Y_2]}{dt}=\dfrac{1}{2}\dfrac{d[XY]}{dt}$ . (2) For $N_2O_5(g) \rightarrow 2NO_2(g)+\dfrac{1}{2}O_2(g)$ , $\text{Rate}=-\dfrac{d[N_2O_5]}{dt}=\dfrac{1}{2}\dfrac{d[NO_2]}{dt}=2\dfrac{d[O_2]}{dt}=2.5\times10^{-3}\ \text{mol dm}^{-3}\text{s}^{-1}$ , so $\dfrac{d[NO_2]}{dt}=5.0\times10^{-3}$ and $\dfrac{d[O_2]}{dt}=1.25\times10^{-3} …

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