I have read in some books that when pressure increases, the number of effective collisions increases. Hence, $k$ depends on pressure. Is it true? If so, then why doesn't $k$ depends on concentration? The number of effective collisions should increase here also.
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2$\begingroup$ In general, equilibrium "constants" depend on temperature, pressure, and other factors. In many circumstances the "constants" are constant enough to provide reasonable values. For examples solutions are a lot less pressure dependent than gases. // If rate = k[A] then as [A] changes so does the rate. Thus there are more or less collisions. $\endgroup$– MaxWCommented Aug 16, 2018 at 19:33
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3$\begingroup$ The rate law already has an explicit dependence on concentration; $k$ is supposed to be the part of the rate law that doesn't depend on concentration. $\endgroup$– a-cyclohexane-moleculeCommented Aug 16, 2018 at 23:18
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$\begingroup$ Depend of pressure? No just as a-cyslohexane-molecule writes. Sometimes there is a confusion between rate and rate constant. Consider a first-order reaction such as isomerisation where only one species is involved. $\endgroup$– porphyrinCommented Dec 7, 2021 at 18:56
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3 Answers
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According to hard-sphere collision theory of gas-phase reactions, the rate constant ($k$) for the elementary bimolecular reaction $\ce{B + C -> products}$ is $$ k = N_\text{A} \pi (r_\text{B} + r_\text{C})^2 \left[ \frac{8RT}{\pi}\left( \frac{1}{M_\text{B}} + \frac{1}{M_\text{C}}\right)\right]^{1/2}e^{-E_\text{thr}/RT}$$ where $N_\text{A}$ is the Avogadro constant, $r_\text{B}$ and $r_\text{C}$ are the radii of the hard-spheres, $R$ is the ideal gas constant, $M_\text{B}$ and $M_\text{C}$ are the molar masses of B and C, and $E_\text{thr}$ is the threshold energy on a per-mole basis. Therefore, no pressure dependence.
For nonideal systems, the rate law for the same elementary reaction can be written as $$ r = -\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = k_r [\text{B}][\text{C}]$$ The Brønsted–Bjerrum equation relates the apparent rate constant, $k_r$ with the infinite-dilution rate constant, $k_r^\infty$, and the activity coefficients of the envolved species: $$ k_r = \left( \frac{\gamma_\text{B}\gamma_\text{C}}{\gamma^\ddagger}\right)k_r^\infty$$ where $\gamma_\text{B}$, $\gamma_\text{C}$, and $\gamma^\ddagger$ are the concentration-scale activity coefficients of B, C, and the transition state respectively. These quantities depend on temperature, pressure, and on the concentration of all species on the same phase. For this reason, the rate constant of nonideal elementary reactions can have a dependence on the pressure.
answered Aug 21, 2018 at 21:42
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According to Arrhenius equation $$k = A \exp\left\{-\frac{E_\mathrm{a}}{R T}\right\},$$ the rate constant $k$ depends on:
- Activation energy $E_\mathrm{a}$: $k$ increases with the decrease of $E_\mathrm{a}$
- Temperature $T$: $k$ increases with the increase of $T$
- Pre-exponential factor $A$: This factor represents the total successfully oriented collisions, so a higher pressure will lead to higher concentration of particles, will lead to a higher the value of $A$, but the value of $A$ changes very little with pressure, we consider it as a constant value, so the effect of pressure on $k$ is very little.
answered Aug 17, 2018 at 12:58
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3$\begingroup$ Is it experimentally observed that A changes very little with pressure? Because I have also seen that in collision frequency expression (theoretical), it is directly proportional to pressure. $\endgroup$– RS2000Commented Aug 19, 2018 at 10:46
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1$\begingroup$ It is just analysis, The total successfully oriented collision that leads to a successful reaction when it has enough energy, and this number very large, so $(A)$ increase with increasing number of collisions, but this effect very little on$ (K)$ in comparison of the pressure effect on increasing successful collisions fraction, so we can neglect the effect of pressure on increasing number of collisions in this discussion . $\endgroup$ Commented Aug 19, 2018 at 11:48
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$\begingroup$ @MartinThank you for your fruitful active effort, I like the formula of the Arrhenius equation as it written in my old textbook $\endgroup$ Commented Aug 21, 2018 at 15:15
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It's so easy. You are to look deep at how actually a book represents the rate of a reaction in term of law of mass action. For a general reaction A + B ........> P
Rate > [A] [B]
Rate = k [A] [B]
K = Rate / [A] [B] .......1
Now look at equation 1. Here k is the ratio of rate of reaction to the product of concentration of reactants. It means when we increase the concentration of reactants the rate of the will increase to the same extent so that to keep the value of k a constant. Same is the case of homogeneous ideal gaseous system where we replace the concentration term by pressure.
