reaction feasibility

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Reaction Feasibility. Will a specific reaction occur ?. How can we predict if a reaction is feasible ?. NaHCO 3 + HCl → NaCl + H 2 O + CO 2. Δ H is +ve (endothermic). Entropy increases. We know this reaction is feasible. How can we predict if a reaction is feasible ?. - PowerPoint PPT Presentation

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Reaction Feasibility

Will a specific

reaction occur ?

How can we predict if a reaction is feasible ?

NaHCO3 + HCl → NaCl + H2O + CO2

We know this reaction is feasible

ΔH is +ve (endothermic) Entropy increases

How can we predict if a reaction is feasible ?

ΔG = ΔH - T ΔS

T = temperature (K)

ΔG = free energy ( kJmol-1)

ΔH = enthalpy change ( kJmol-1)

ΔS = entropy change ( kJK-1mol-1)

ΔS must be converted from JK-1mol-1 (/1000)

How can we predict if a reaction is feasible ?

ΔG = ΔH - T ΔS

A reaction is feasible if ΔG is negative

ΔG < 0

A reaction becomes feasible when

ΔG = 0

Calculate ΔG for the formation of CO2 from graphite at 298K, given ΔHf = -393.5kJmol-1

& ΔS is +3.3 JK-1mol-1

ΔG = ΔH - T ΔS

(-393.5) – (298 x 3.3/1000)

ΔG = -394.5 kJmol-1

Feasible reaction ΔG = -ve

Is the decomposition of limestone spontaneous at standard temperature? ΔS = +165 JK-1mol-1 ΔH = +178 kJmol-1

ΔG = ΔH - T ΔS

(+178) – (298 x 165/1000)

ΔG = +128.83 kJmol-1

Not feasible ΔG = + ve

Feasible Spontaneous

Thermodynamics tells us nothing about the speed a reaction happens

It only tells us that it can happen

C + O2 → CO2 ΔG = -394.5 kJmol-1

EA is large so reaction very slow at 298K

Calculate ΔG for the rusting of Fe at 298K, given ΔHf = -825 kJmol-1

& ΔS is -272 JK-1mol-1

ΔG = ΔH - T ΔS

(-825) – (298 x (-272/1000)

ΔG = -744 kJmol-1

Feasible reaction ΔG = -ve

2Fe(s) + 3/2O2(g) → Fe2O3 (s)

At what temperature does this reaction cease to be feasible?

ΔHf = -825 kJmol-1 & ΔS is -272 JK-1mol-1

ΔG = ΔH - T ΔS Reaction becomes feasible when ΔG = 0

ΔH = T ΔS

T = -825/-0.272

ΔH = T ΔS

= 3033K

Free energy & Equilibrium

ΔG = 0 for a system at equilibrium

Changes of state are systems at equilibria

ΔH = T ΔS ΔH = T ΔS

ΔS =ΔH T

ΔG = ΔH - T ΔS

Calculate the entropy change when ice melts given that ΔHfusion (melting) = + 6.0 kJmol-1

ΔH = T ΔS

ΔS = + 6/273

= 0.0219 kJK-1mol-1

ΔS =ΔH

T

ΔG = 0 for a change of state ΔG = ΔH - T ΔS

= 0.0219 x 1000 JK-1mol-1 = 21.9 JK-1mol-1

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