Effect of Concentration on Rate; Rate Expressions

CHERate, Order & Molecularity

Here the rate law is given (or extracted from data) and you predict how the rate responds when concentrations change. The work is arithmetic with the exponents in the rate expression and reading orders off initial-rate tables.

Rate law and response factor

\text{rate} = k[A]^x[B]^y \quad\Rightarrow\quad \dfrac{\text{rate}_2}{\text{rate}_1} = f_A^{\,x}\,f_B^{\,y}

f_A,f_B

= factors by which

[A],[B]

change

Order by logarithms

x = \dfrac{\log(\text{rate}_2/\text{rate}_1)}{\log([A]_2/[A]_1)}

[B]

held constant; overall order

n=x+y

Rate constant from one run

k = \dfrac{\text{rate}}{[A]^x[B]^y}

substitute any single experiment's data

From $\text{rate} = k[A]^x[B]^y$ , multiplying $[A]$ by a factor $f$ (others fixed) multiplies the rate by $f^x$ .
Example: $\text{rate} = k[A]^2[B]$ — doubling $[A]$ alone makes the rate $2^2 = 4$ times faster.
Example: $\text{rate} = k[A][B]^2$ — doubling both $[A]$ and $[B]$ changes the rate by $2^1 \times 2^2 = 8$ times.
Find order in each reactant by the ratio method: take two experiments where only that reactant's concentration differs and compare the rates.
Overall order $= x + y$ ; once orders are known, compute $k$ by substituting any one experiment's concentrations and rate into the rate law.
Use logarithms when concentration ratios are not simple: order $x = \dfrac{\log(\text{rate}_2/\text{rate}_1)}{\log([A]_2/[A]_1)}$ at fixed $[B]$ .
Always derive the rate law from experimental data, not the stoichiometric coefficients of the balanced equation.
If changing a reactant's concentration leaves the rate unchanged, the order in that reactant is zero ( $f^0=1$ ).
Halving a concentration with order $x$ multiplies the rate by $(1/2)^x$ ; the same factor logic works for any fractional change.
When both concentrations change, multiply the individual response factors: total factor $=f_A^{\,x}\times f_B^{\,y}$ .
After finding all orders, verify $k$ by computing it from a second experiment — consistent $k$ confirms the rate law.
Quote $k$ with the correct units for the overall order found, e.g. second order gives $k$ in $L\,mol^{-1}s^{-1}$ .

Where the marks go

Using stoichiometric coefficients as the exponents in the rate law instead of experimentally determined orders.
Forgetting to hold the other reactant constant when extracting one reactant's order from a data table.
Multiplying the factor by the order instead of raising it to the power of the order — doubling $[A]$ with order $2$ makes the rate $2^2=4$ times faster, not $2\times 2$ added on.
Treating a zero-order reactant as if it still affected the rate when its concentration changes.
Reporting $k$ without units, or with units that contradict the overall order just determined.

How the board asks it

Numericalorder by the ratio method from an initial-rate table
For the reaction $A + B \rightarrow$ products, the following initial-rate data were recorded in three experiments in which $[A]$ and $[B]$ are varied and the initial rate is measured. Determine the order with respect to $A$ , the order with respect to $B$ , the overall order, and calculate the value of the rate constant $k$ with its units.
Predict the productresponse factor $f^x$ applied to $\text{rate}=k[A]^x[B]^y$
The rate law for a reaction is $\text{rate} = k[A]^2[B]$ . By what factor does the rate change if $[A]$ is doubled and $[B]$ is halved simultaneously?
Numericalorder from logarithms when the concentration ratio is not a simple whole number
When $[A]$ is increased from $0.10\,mol\,L^{-1}$ to $0.15\,mol\,L^{-1}$ at fixed $[B]$ , the initial rate increases from $4.0 \times 10^{-3}$ to $9.0 \times 10^{-3}\,mol\,L^{-1}\,s^{-1}$ . Using logarithms, calculate the order of the reaction with respect to $A$ .
Define / stateoverall order $=x+y$ and units of $k$
Write the rate-law expression for a reaction that is first order in $A$ and second order in $B$ . State its overall order and give the units of the rate constant $k$ .
Give reasonsorder is determined experimentally, not from stoichiometric coefficients
For the reaction $2NO + 2H_2 \rightarrow N_2 + 2H_2O$ , the experimentally determined order with respect to $H_2$ is $1$ , although its stoichiometric coefficient is $2$ . Explain why the order of a reaction with respect to a reactant need not equal its stoichiometric coefficient in the balanced equation.
Assertion–Reasona zero-order reactant leaves the rate unchanged ( $f^0=1$ )
Assertion: When the concentration of a particular reactant is doubled, the rate of the reaction remains unchanged. Reason: The order of the reaction with respect to that reactant is zero. State whether both statements are true and whether the Reason correctly explains the Assertion.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.