June 2024
Remember to ALWAYS use ICE table to simplify the reaction stoichiometry, concentrations, find degree of dissociation and such quantities in chemical kinetics AND chemical and ionic equilibrium.
For a reaction, the rate in which a product is formed, or the reactants are consumed is generally proportional to some power of the concentration of the reactant. Take a reaction
\[\ce{A -> B}\]
Here, the rate of the reaction is proportional to \([\ce A]^n\), so
\[\text{rate} = k[\ce A]^n\]
The average rate of the reaction is the macroscopic change in concentration of reactant (\(\Delta c\)) or product over the time it took for that change, as
\[\text{rate}=\frac{\Delta c}{\Delta t}\]
The instantaneous rate is the derivative of the concentration w.r.t time, as
\[\text{rate}_\text{inst}=\frac{dc}{dt}\]
For a reaction
\[\ce{aA + bB -> cC + dD}\]
The rate of the depletion of A by \(a\) is equal to that of B by \(b\) and appearance of C by \(c\) and D by \(d\). Then, the rate of the reaction is
\[\text{rate of reaction} = -\frac{d[\ce A]}{dt}\frac1a=-\frac{d[\ce B]}{dt}\frac1b=\frac{d[\ce C]}{dt}\frac1c=\frac{d[\ce D]}{dt}\frac1d\]
For a homogeneous gas-phase reaction, the rate of the reaction can be expressed in terms pressure. From \(pV=nRT\), we get \(p=cRT\), then \(c=p/RT\). This yields
\[\text{rate of reaction} = -\frac{dp_\ce A}{dt}\frac1a=-\frac{dp_\ce B}{dt}\frac1b=\frac{dp_\ce C}{dt}\frac1c=\frac{dp_\ce D}{dt}\frac1d\]
It states that
the rate of a reaction is directly proportional to the active mass of reactants raised to the power of their stoichiometric coefficients.
For a reaction
\[\ce{aA(g) + bB(g) -> cC(g) + dD(g)}\]
\[\text{rate}=k[\ce A]^a[\ce B]^b\]
This rate is the theoretical rate.
In certain reactions only is the experimental rate equal to the theoretical rate.
It is based on the law of mass action, but instead of the stoichiometric coefficients, substitute experimentally determined variables instead, as
\[\text{rate}_E=k[\ce A]^x[\ce B]^y\]
In reactions where theoretical rate is the same as the experimental rate, both \(x,y\) and \(a,b\) become equal.
The rate constant \(k\) of a reaction is the rate of the reaction when the concentration of all the reactants is unity (1).
Characteristics of the rate constant:
The rate constant can never decrease with increase in temperature.
It is a theoretical concept, the total number of reacting species which must collide simultaneously to give products.
For example:
It can
While complex reactions by themselves cannot have a molecularity, their individual steps can. For example, in the reaction given in the section classification of reactions, the individual steps have molecularity 2 and 2 respectively.
Importantly, for finding molecularity, the stoichiometric coefficients of all reactants and products must be a natural number, i.e., for \(\ce{1/2 H2 + 1/2 I2 -> HI}\), multiply the entire reaction twice to get molecularity 2.
The order of the reaction is the sum of the powers of the concentration terms involved in the rate law expression.
For a reaction
\[\ce{a A + b B -> products}\]
Let the rate law expression be
\[r=k[\ce A]^x[\ce B]^y\]
Then, the order of the reaction w.r.t A is \(x\) and B is \(y\). The overall order of the reaction is \(x+y\). This must always be greater than 0, while the individual orders can be negative.
The order of a reaction tells us how much the rate of a reaction changes on changing the concentration of the reactants.
The reactions where the rate of the reaction is completely independent of the concentration of the reactants are called zero order reactions.
For a reaction
\[\ce{A -> B}\]
The rate law expression is
\[r=k\]
Order of reaction is 0, so \(r=k[\ce A]^x\) becomes \(r=k[\ce A]^0\). From this, it can be found that \(k=-d[\ce A]/dt\), then multiply both sides by a \(dt\) and integrate to get \([\ce A]^{A_t}_{0}=-kt\). This solves to give
\[A_t=A_0-kt\label{eq1}\tag{i}\]
The half life (\(t_{0.5}\) or \(t_{1/2}\)) is the time taken for half of the reaction to complete, i.e., for the concentration of the reactant to become half of its initial concentration. For \(A_t=A_0\), it can be solved from (\(\ref{eq1}\)) that
\[t_{1/2}=\frac{A_0}{2k}\]
The time taken to complete 100% of a reaction. This is easily found to be \(2t_{1/2}\) for a zero order reaction, or
\[t_{1}=2t_{0.5}=\frac{A_0}{k}\]
The degree of dissociation \(\alpha\) is the number of moles dissociated (\(A_0-A_t\) in 1L) over the total number of moles (\(A_0\)). Substituting \(A_t\) for (\(\ref{eq1}\)), we get
\[\alpha =\frac{kt}{A_0}\]
The unit of the rate constant for a zero order reaction is \(\mathrm{mol~L^{-1}~s^{-1}}\).
NOTE: If in equal time intervals, the change in concentration of reactants remains the same, then it is a zero order reaction.
rate vs. concentration is a straight line with slope 0, since it is constant.
A reaction where the rate depends linearly on the concentration of the reactant is called a first order reaction.
For a reaction
\[\ce{A -> B}, x=1\]
Let \(A_0\) be the initial concentration of the reactant and \(A_t\) be the final concentration. The rate law expression for this reaction is
\[k[\ce A] = -\frac{d[\ce A]}{dt} \implies k\int dt=-\int\frac{d[\ce A]}{[\ce A]}\]
This equation solves to give
\[\ln A_t = \ln A_0 - kt\label{eq2}\tag{ii}\]
Graphing \(\ln A_t\) against \(t\) gives a straight line with slope \(k\) (negative) and \(y\)-intercept \(\ln A_0\) as given below.
Convert the equation to base 10 logarithm using 2.303. Also, solve to get
\[k=\frac{1}{t}\ln\left(\frac{A_0}{A_t}\right)\quad \text{OR} \quad k=\frac{2.303}{t}\log_{10}\left(\frac{A_0}{A_t}\right) \label{eq3}\tag{iii}\]
This also yields
\[\frac{A_t}{A_0}=e^{-kt}\]
I’ve already graphed equation (\(\ref{eq2}\)) in this section.
From the equation (\(\ref{eq3}\)), we may substitute \(A_t=A_0/2\) to get
\[t_{1/2}=\frac{\ln 2}{k}\implies t_{1/2}=\frac{0.693}{k}\]
This equation tells us that the half life of a first order reaction is independent of the initial concentration of reactant. It is only dependent on the rate of the reaction.
From the equation (\(\ref{eq3}\)), we may substitute \(A_t=0\) which gives a 0 as the denominator, and as such
\[t_1=\infty\]
The reaction will never complete, therefore it is a reversible reaction, which has its own equilibrium constant (\(K_c\)).
Drawing the ICE diagram for the reaction \(\ce{A -> B}\),
| time | \(\ce{A}\qquad \to\) | \(\ce{B}\) |
|---|---|---|
| \(t=0\) | \(A_0\) | \(0\) |
| \(t=t\) | \(A_t\) | \(B_t\) |
We know that \(\alpha = (A_0-A_t)/A_0\). For \(A_t=A_0e^{-kt}\), we get
\[\alpha = 1-e^{-kt}\]
The degree of dissociation is independent of initial concentration of reactant.
The number of half lives is \(t/t_{0.5}\), so then the amount of reactant left after \(n\) half lives becomes
\[A_{\text{unused}}=\frac{A_0}{2^n}\]
For a zero order reaction, for the same time intervals, the difference in concentrations is always equal. In first order reactions, it is the final over initial concentration that is constant, as
\[\frac{A_2}{A_1}=\frac{A_4}{A_3}\]
From \(pV=nRT\), it is known that \(p\propto n\). For a reaction \(\ce{A(g) -> B(g) + C(g)}\), drawing the ICE table,
| time | \(\ce{A(g)}\qquad \to\) | \(\ce{B(g)}\qquad +\) | \(\ce{C(g)}\) |
|---|---|---|---|
| \(t=0\) | \(p_0\) | \(0\) | \(0\) |
| \(t\) | \(P_0-x\) | \(x\) | \(x\) |
From Dalton’s Law, we state that \(p_T=\sum p\). For \(t=0\) then, we have \(p_T=p_0\). At \(t=\infty\), we have \(p_T=2p_0\), which we can rewrite as \(p_\infty=2p_0\) (since A gives B and C with same stoichiometric coefficients).
Since \(p\propto n\), and concentration is basically the number of moles per unit substance (volume, mass, etc.), we substitute the concentration terms with pressure terms, then equation (\(\ref{eq3}\)) becomes
\[k=\frac{1}{t}\ln\left(\frac{p_0}{p_0-x}\right)\]
To remove the \(x\), we look back at the ICE table, where we find that \(p_T\) at \(t\) is \(p_T=p_0-x+x+x=p_0+x\). This allows us to find \(x\) as \(p_T-p_0\). Substitute to get
\[k=\frac1t\ln\left(\frac{p_0}{2p_0-p_T}\right)\]
In terms of \(p_\infty\) and \(p_T\), the equation becomes
\[k=\frac1t\ln\left(\frac{p_\infty}{2[p_\infty-p_T]}\right)\]
The reactions in which the reactant is involved in more than one first order reaction are called parallel first order reactions.
In parallel first order reactions, the average rate constant
\[k_\text{avg}=k_1+k_2\]
If there are more than two rate constants, then just add them as well. This new constant can be used in all equations that are followed by first order reactions. For example, we may use equation (\(\ref{eq3}\)) here as:
\[k_1+k_2=\frac{1}t\ln\left(\frac{A_0}{A_t}\right)\]
This would also solve to become
\[A_t=A_0e^{-(k_1+k_2)t}\]
Take the reaction given in the above section.
Here, the percentage of B formed is
\[\%\ce B=\frac{k_1}{k_1+k_2}\times 100\]
The percentage of C formed is
\[\%\ce C=\frac{k_2}{k_1+k_2}\times 100\]
Take the reaction given in the above section.
Here, the number of moles of B formed is
\[n_\ce B = \frac{k_1}{k_1+k_2}\times x\]
where \(x\) is \(A_0-A_t\) of the reactant.
The number of moles of C formed is
\[n_\ce C = \frac{k_2}{k_1+k_2}\times x\]
Since \(t_{0.5}=0.693/k\), for \(k=k_1+k_2\) we have
\[t_{0.5}=t_{{0.5}_1}+t_{{0.5}_2}\]
For a zero-order reaction, \(t_{0.5}\propto A_0\), and for a first order reaction it is not dependent. It can then be proven that for a reaction
\[\ce{A -> B}\]
where \(A_1\) is the concentration of reactant at \(t_{{0.5}_1}\) and \(A_2\) at \(t_{{0.5}_2}\),
\[\frac{t_{{0.5}_1}}{t_{{0.5}_2}}=\left(\frac{A_1}{A_2}\right)^{1-n}\]
where \(n\) is the order of the reaction.
The reaction which proceeds from reactants to final products through intermediate stages.
\[\ce{A ->[$k_1$] B ->[$k_2$] C}\]
Rates for each step differ as \(k[\ce A]\), \(k[\ce B]\) and so on.
The reaction whose order is different from the actual order due to large concentration of one of the reactants is called a pseudo order reaction.
\[\ce{A + B -> Products}\]
Then, rate \(k[\ce A][\ce B]\), when B is in excess becomes \(k'[\ce A]\).
Examples
These above are pseudo first order reactions.
this is all i have written at the moment.