# Physical Chemistry – Kinetic Theory of Gases

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A gas:

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A fluid state of matter that fills any container, is easily compressed, mixes with any gas, and exerts pressure on its surroundings.

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Properties:

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Static properties – pressure p, volume v, temperature T Transport – diffusion/effusion, viscosity Thermal – heat capacity Chemical – reactions between gases

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Boyle’s Law

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*[ V α 1/p ]* At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

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Charles’ Law

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*[ V α T ]* At constant pressure, the volume of a fixed mass of gas is directly proportional to its temperature measured on the Kelvin scale. Implies minimum temperature (absolute 0) OR *[ p α T ]* At constant volume, p=constant x T i.e. p->0 as T->0 (limiting law)

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Combining Results:

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pV = constant V = constant x T p = constant x T All molar volumes of gases should be the same: Vm = V/n or V = constant x n pV/T = constant => pV = constant x T

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Value of constant:

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Under standard conditions defined as S.T.P. P = 1 atm = 101 325 Pa T = 273.16 K (triple point of water) Volume of 1 mole of an ideal gas = 22.414 dm^3 constant = pV/nT = (101325 x 22.414)/(1 x 273.16) = 8.31441 J/(K mol) = Gas constant, R *[ Ideal gas equation => pV = nRT ]* Boltzmann constant, k = R/NA (per molecule)

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Assumptions of the ‘Kinetic Theory of Gases’

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1) Gas consists of molecules of mass m and diameter d in ceaseless, random motion 2) Volume occupied by molecules is negligible compared to the volume of the gas 3) Molecules do not interact with one another, with the exception of perfectly elastic collisions (overall energy conserved)

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Predicting Pressure

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*Pressure = Force/Area* *Force = Rate of change of momentum = m(v-u)/t*

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Momentum

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Momentum = mv (vector quantity) Imagine a molecule travelling at a velocity v colliding with a wall (elastic collision) v = [(vx)^2 + (vy)^2]^(1/2) vx = horizontal component vy = vertical component before collision: component towards wall (horizontal) Momentum = m(vx) after collision: component towards wall Momentim = -m(vx) [moving away from the wall with the same speed as energy is conserved] *Change in momentum (molecule) = -m(vx)-m(vx) = -2m(vx)* *Change in momentum (moles) = -2M(vx)*

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Molecules Hitting the Wall

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In a given time Δt a molecule travels a distance vxΔt along the x axis. So, all molecules at a distance vxΔt or less travelling towards the wall will reach the wall in Δt In 3D: For a unit area A of wall, molecules have to be in a ‘collision volume’ of vxΔtA. *number of molecules per unit volume = (n x Na)/V* *number in ‘collision volume’: (vx.Δt.A.n.Na)/V* approximately 50% of the molecules are travelling towards the wall, therefore: *[ number of molecules colliding with the wall in a time Δt = (vx.Δt.A.n.Na)/2V ]*

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Change in Momentum

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Total change in momentum = number of collisions x Δ(mv)/collision = (vx.Δt.A.n.Na)/2V x 2M(vx) [note m.Na = M] Rate of change of momentum = total change/Δt = (n.M.A.(vx)^2)/V = force pressure = force/area *[ Pressure = (n.M.(vx)^2)/V ]*

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Average

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Molecules have a range of vx Pressure should depend on average (vx)^2 => Random motion of molecules i.e. no net flow in one direction => = = Mean square speed of molecules c^2 = + + = 3 = (c^2)/3 Thus, our model predicts: *[ p = (m.M.c^2)/(3V) or pV = (1/3)(n.M.c^2) ]*

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Comparing Model with Experimental results

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Experimental: pV = nRT Model: pV = (1/3)(n.M.c^2) Thus, if c^2 depends only on Tm then at constant T model predicts pV = constant Model also implies RT = (1/3)M.c^2 Hence, mean square speed, *c^2 = (3RT)/M* => root mean square (r.m.s.) speed: *c=(3RT/M)^(1/2) = (3kT/m)^(1/2)*

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Kinetic Energy

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For one molecule, kinetic energy = (1/2)mv^2 Total kinetic energy = N.(1/2)mv^2 Model predicts -> pV = (1/3)nMc^2 (nM=Nm) so, pV =(1/3)Nmc^2 this implies that pV = (2/3)(k.e.) or k.e.= (3/2)pV For our ideal gas, pV=nRT, therefore: *[ k.e. = (3/2)nRT ]* Translational energy is the only energy for our ideal model gas As c^2 = + + k.e. = N(1/2)M(vx)^2 + N(1/2)M(vy)^2 + N(1/2)M(vz)^2 as (vx)^2 = (vy)^2 = (vz)^2 Total k.e. = (3/2)nRT k.e for each translational degree of freedom = (1/2)nRT

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Molecular Speeds

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Molecules collide many times in their constant motion, therefore molecular speeds are wide ranging. We show the range of speeds as a *probability density/distribution function*

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Maxwell Distribution

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probability distribution funtion showing range of speeds of molecules, the Maxwell distribution: *[ f(s) = 4π(m/2πkT)^(3/2).(s^2)exp(-ms^2/2kT) ]* distribution not symmetrical peak of f(s) -> most probable speed, s* Experimental verification of the Maxwell distribution: experiments use a velocity selector – only molecules with specific speeds make it through the slots of the rotating discs to the detector As T is increased, peak value of s increases and probability decreases As M is increased, peak value of s decreases and probability increases

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Most Probable Speed, s*

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As Maxwell distribution is asymmetric, most probable speed, s* is less than mean speed, ~~ which is less that mean square speed, ~~~~ s* < ~~~~ < ~~~~ s* can be found by differentiating f(s) and setting it equal to 0 i.e. df(s)/ds = 0 Use product rule f'(s) = 4π(m/2πkT)^(3/2)[(s^2)(-ms/2kT).exp(-ms/2kT) + 2s.exp (-ms/2kT)]~~

Mean Speed, ~~ and Mean Square Speed ~~

Molecular Collisions

The Maxwell distrivution shows that ~~ = (8kT/πm)^(1/2) Considering collisions we need relative mean speed, srel = ~~~~rel Relative speeds? Two extremes: -two molecules travelling in the same direction with the same ~~~~ => ~~~~rel = 0 -two molecules travelling in opposite directions with the same ~~~~ => ~~~~rel = 2~~~~ For everything in between, need to consider the mean direction of approach of molecules Mean speed of approach: resolve molecular (speed) velocity into two components (a) => (a^2) + (a^2) = ~~~~^2 2(a^2) = ~~~~^2 a = ~~~~/√2 therefore, component of mean speed towards other molecule = ~~~~/√2 hence, mean relative speed = 2~~~~/√2 = √2/~~

Reduced Mass

if molecules have different masses, we have to use reduced mass, μ: *μ = mAmB/(mA+mB)* if mA = mB μ=(mA^2)/2 ~~ = (8kT/πm)^(1/2) ~~~~rel = √2(8kT/πm)^(1/2) ~~~~rel = (8kT/πμ)^(1/2)~~

Collisions Between Molecules

Imagine all molecules, except one are static. Give the moving molecule a speed ~~rel. So, in a time Δt, our molecule effectively travels a distance ~~~~rel.Δt The molecule will hit any other molecule when the centre separation of the molecules is d (diameter of the molecule) The molecule thus sweeps out a ‘collision tube’ oh diameter 2d (radius d) and length ~~~~rel.Δt The molecule will hit any other molecule in the collision tube Cross-sectional area of tube = πd^2 = σ Volume of tube = σl = σ.~~~~rel.Δt Number of collisions made by one molecule in time Δt = σ.~~~~rel.Δt. (N/V) Number of collisions made by one molecule per unit time = σ.~~~~rel.(N/V) *[ ZA = σ.√2~~~~.(N/V) = σ.√2~~~~.(P/kT) ]* {pV = NkT} Mean Free Path If a molecule collides at a frequency ZA Time between collisions = 1/ZA Distance between collisions = ~~~~/ZA = λ (mean free path) Total number of collisions= ZA(NA/2V) {divide by 2 to avoid double counting} *[ ZAA = ZA(1/2kT) ]*~~

Collision Between Different Molecules

Collisions for one molecule of A in gas B Collision cross-section, σAB = π((da +db)/2)^2 ZB = σAB.~~rel.(N/V) *[ ZAB = ZB(Na/2V) ]* {number of collisions between A and B molecules per unit time per unit volume}~~

Applications of Collision Frequency

Collision frequency is used in developing a theory to explain kinetics (rates) of reactions. Why are kinetics important? – Allows us to understand/test reaction mechanisms – Allows us to apply reactions e.g environentally, industrually

Gas Phase Reaction Kinetics

For a reaction: aA + bB -> cC + dD The rate of reaction is * [ r = (-1/a).(d[A]/dt) = (-1/b).(d[B]/dt) = (+1/a).(d[C]/dt) = (+1/d).(d[D]/dt) ]* Experimentally: *[ r = k.[A]^α.[B]^β ]* k: rate constant α: orders of reaction for A β: orders of reaction for B {Experimentally determined} Overall order of reaction: α + β If a reaction is elementary (single step), order of reaction = molecularity (number of molecules taking part) A -> products (unimolecular) A + B -> products (bimolecular) A + B +C -> products (termolecular) An elementary bimolecular reaction in the gas phase can be considered as a collision process.

Collision Theory for Reaction Rates

Two assumptions made: 1) The reactants must collide 2) On collision the reactants must have enough energy to react (so, for A + B -> products) *[ Theoretical rate reaction = collision frequency (ZAB) x Fraction of molecules with enough energy to react ]*

The Boltzmann Distrbution

To find the fraction with enough energy to react we must look at the probability distribution function for energies i.e. the Boltzmann distribution *f(E) = (1/kT)exp(-E/kT)* Look at area under the curve above E = ε, where ε is the minimum energy needed for the molecules to react. => ∫ f(E) dE (limits ε to +∞) = exp(-ε/kT)

Theoretical and Experimental Rate of Reaction

Theoretical rate = ZAB x exp(-ε/kT) = σAB.~~rel.(NB/V).(NA/2V).exp(-ε/kT) {NA and NB are number of molecules} = (σAB.~~~~rel.[B][A].NA^2)(exp(-ε/kT))/2 {NA = Avagadro’s number} Experimental rate = k[A][B] Hence, theory predicts that: *[ k = (σAB.~~~~rel.NA^2)(exp(-ε/kT))/2 ]*~~

Experimental Rate Constant vs. T

Many reactions show a rate constant that follows the Arrheinius equations: *[ A.exp(Ea/RT) ]* {A = pre-exponential factor, Ea = activation energy} Theoretical k = (σAB.~~rel.NA^2)(exp(-ε/kT))/2 therefore *[ A = σAB.~~~~rel.NA^2/2 ]* Steric Factors: Collision theory usually overestimates the experimental reaction rate, because molecules do not always react on collisions (as molecules must be correctly oriented) The steric factor, P is defined as *[ P = Aexperimental/Atheoretical ]* usually, P<1 as Aexperimental<Atheoretical~~

Effusion

The emergence of gas from a small hole in a container. Treat this in the same way as ‘how many molecules hit a given area of wall’ In a time Δt, molecules travel a distance vxΔt in the direction towards the wall. So for an area A of wall, molecules in a collision tube of vxΔt hit the wall . We need to use the mean vx, going towards the wall Use a probability distribution function f(vx) = ∫vx.f(vx) dvx (limits -∞ to +∞) f(vx) = (m/2πkT)^(1/2).exp(-m(vx)^2/2kT) {this differs from the speed distribution} vx can be + or – only count positive vx = (kT/2πm)^1/2 Number of molecules hitting the wall: *Zw = ΔtA(N/V)/AΔt = (N/V) = (N/V)(kT/2πm)^(1/2) = (p/kT)(kT/2πm)^(1/2) = p/(kT2πm)^(1/2)* *[ Zw = (N/V) x (~~rel/4) = (p/kT) x (~~~~rel/4) ]*~~

Graham’s Law of Effusion

Graham’s law states that the rate of effusion of a gas from a small hole is inversely proportional to (molar mass)^(1/2) If the wall has hole in it of area A0: *Rate of effusion = Zw.A0 =p.A0/(kT2πm)^(1/2) =p.A0.L/(MRT2π)^(1/2)* {m = M/L, k = R/L} Knudsen applied Graham’s law to measure (very low) vapour pressures of solids/liquids by measuring loss of mass, Δm by effusion in a time Δt *[ Δm = Zw.A0.m.Δt ]* {Zw.A0 = rate of loss of molecules, m = molecular mass} Zw = Δm/(A0.m.Δt) = p/(2kTπm)^(1/2) {rearrange} p = (Δm/A0.m.Δt) (2kTπm)^(1/2) *[ p = (Δm/A0.m.Δt) (2kTπ/m)^(1/2) ]* *[ p = (Δm/A0.m.Δt) (2RTπ/M)^(1/2) ]*

Real Gases and the Compression Factor, Z

Real gases deviate from perfect behaviour Z = p(Vm)/RT {Vm = molar volume, Vm = V/n} For an ideal gas, Z = 1 low pressures – real gases behave like ideal gases i.e Z~1 moderate pressures – *long range attractive forces* draw molecules together, hence *Z1* Z also depends on mean speeds of molecules low T (or high M) slow moving molecules are more readily captured by one another i.e. Z1 At the Boyle temperature, TB (an intermediate temperature), a real gas behaves ideally over a range of p

The Virial Coefficients

At large Vm and high T (low p) gas behaves more ideally At small Vm and low T (high p) behaviour is non-ideal So, P(Vm) = RT can be expanded to account for non-ideality: *[ p(Vm) = RT(1 + B’p + C’p^2) ]* *[ p(Vm) = RT(1+B/(Vm) + C/(Vm)^2) ]* There are versions of the virial equations p(Vm) = RT(1+B/(Vm) + C/(Vm)^2) +… {1st virial coefficient = 1, 2nd virial coefficient = B, 3rd virial coefficient = C} Virial coefficients are dependent on T At typical molar volumes B/(Vm) >> C/(Vm)^2) Virial equations predict Z = p(Vm)/RT = 1 when (Vm) -> ∞ (p -> 0) The slope of Z with p (or 1/(Vm)) differs in virial equations *For an ideal gas: Z = 1, dZ/dp = 0 For a real gas Z = 1 + B’p + C’p^2+…, dZ/dp = B’ + 2C’P Z= 1+B/(Vm) + C/(Vm)^2) +…, dZ/dp = B + 2C/(Vm) = B as (Vm) -> ∞* The slope of Z with p or 1/(Vm) is 0 when B=0 at the Boyle temperature, TB, B(TB)=0

Condensation

At low temperatures (below Tc) increasing the p of a gas can lead to liquefaction or even solidification {Tc is the critical T (critical isotherm): Lowest T at which liquid phase can form by compression} above Tc a fluid which fills the entire volume exists, often the fluid is denser than a gas. The critical pressure (pc), volume (vc) and temperature (Tc) are constants for a given gas.

The van der Waals equation

A more realistic model of the gas based on p(Vm) = RT It takes into account: 1) The volume of the gas molecules themselves Volume = Vm – b hence p = RT/(Vm-b) 2) The reduction of the pressure due to: attractive forces between the molecules reducing the *frequency* and *force* of collisions, both α (1/Vm) Hence reduction in p is a/(Vm^2) Thus *[ p = RT/(Vm-b) – a/(Vm^2) ]* or for n moles of gas *[ mRT/(V-nb) – an^2/v^2 ]* as {Vm = V/n} a and b depend only on the nature of the gas, these can be linked to virial coefficients B and C

Molecular Interactions

Look at potential energy of interaction U(R) as a function of molecular separation, R For the ‘hard sphere’ model, p.e. = 0 until molecules collide. The model molecules are impenetrable spheres, the p.e. of interaction is ∞ at R < 0 where σ = d (in this case)

Improvements to the Potential

The simplest improvement is the square well potential. The model still uses hard spheres, but allows the total p.e to be lowered by ε at intermediate values of R. This model accounts for the complex formation , and the existence of condensed phases (liquids and solids)

Realistic Potential Energy of Interactions

We must consider the nature of the forces between molecules – Repulsive forces: when molecules are close together the electrons and nuclei of the different molecules repel one another (+ deshielding). Hence, the potential energy goes up with smaller R: * [ Urepulsion(R) = Aexp(-BR) ]* {A and B are constants} – Attractive Forces: At large separation molecules attract one another. Polar molecules do this by Coulombic attraction. Non-polar molecules do so because of: 1) e- motion => instantaneous dipole 2) the molecule polarises a second molecule and there is an attraction 3) induced dipole The attractive interaction between molecules is known as a dispersion interaction or a London interaction. This depends on: 1) The polarisability of molecules (α) 2) The extent of control of the nucleus over the outer electrons (Ionisation energy) (I) 3) And most importantly – on separation (R) Hence, we write: *[ Urepulsion(R) = -C6/R^6 ]* This is the London formula

The Total Interaction

The overall effect of the attractive and repulsive potentials is obtained as the sum. As an approximation the exponential repulsive part is replaced by R^-12 Hence: *[ Urepulsiion(R) ~ C12/R^12 ]* Thus: *[ Utotal(R) = C12/R^12 – C6/R^6 ]* {C6 and C12 are constants}

The Lennard-Jones Potential

A more realistic potential energy to interactions of two molecules: *[ U(R) – 4ε(σ^12/R^12 – σ^6-R^6) ]* {ε = well depth – minimum p.e. of interaction σ = separation when U(R) = 0 i.e. V(σ) = 0} σ0 = zero potential separation, separation at minimum potential ε can be calculated, because at the minimum: dU(R)/dR = 0 The force between molecules, F = -dU(R)/dR hence, at ε, F=0 general expression for force: *[ F = (24ε/σ)[2(σ/R)^(13) – (σ/R)^(7)] ]* {from differentiations} The attractive force maximises at R = (26/7)^(1/6)σ The Lennard-Jones parameters allow calculation of virial coefficients, hence non-ideality can be predicted This is often called the Lennard Jones 12-6 potential