Part of PC-04 — Chemical Thermodynamics

Formula Sheet (LaTeX)

by Notetube Official216 words7 views

Core Equations

ΔU=q+w(First Law, IUPAC)\boxed{\Delta U = q + w} \quad \text{(First Law, IUPAC)}

ΔH=HproductsHreactants=ΔU+ΔngRT\boxed{\Delta H = H_{products} - H_{reactants} = \Delta U + \Delta n_g RT}

ΔG=ΔHTΔS(Gibbs equation)\boxed{\Delta G = \Delta H - T\Delta S} \quad \text{(Gibbs equation)}

ΔG=RTlnK=2.303RTlogK\boxed{\Delta G^\circ = -RT\ln K = -2.303\,RT\log K}

ΔS=qrevT\boxed{\Delta S = \frac{q_{rev}}{T}}

Work Expressions

wfree=0(Pext=0)w_{free} = 0 \quad (P_{ext} = 0)

wirrev=Pext(V2V1)=PextΔVw_{irrev} = -P_{ext}(V_2 - V_1) = -P_{ext}\,\Delta V

wrev=nRTlnV2V1=2.303nRTlogV2V1w_{rev} = -nRT\ln\frac{V_2}{V_1} = -2.303\,nRT\log\frac{V_2}{V_1}

Enthalpy Relations

ΔHrxn=ΔHf(products)ΔHf(reactants)(Hess’s Law)\Delta H_{rxn}^\circ = \sum\Delta H_f^\circ(\text{products}) - \sum\Delta H_f^\circ(\text{reactants}) \quad \text{(Hess's Law)}

ΔHbond=BE(bonds broken)BE(bonds formed)\Delta H_{bond} = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})

Heat Capacities

CpCv=R(any ideal gas)C_p - C_v = R \quad \text{(any ideal gas)}

qv=nCvΔT=ΔU(isochoric)q_v = nC_v\Delta T = \Delta U \quad \text{(isochoric)}

qp=nCpΔT=ΔH(isobaric)q_p = nC_p\Delta T = \Delta H \quad \text{(isobaric)}

Heat Capacity Values

GasCvC_vCpC_pγ\gamma
Monoatomic3R2\frac{3R}{2}5R2\frac{5R}{2}53\frac{5}{3}
Diatomic5R2\frac{5R}{2}7R2\frac{7R}{2}75\frac{7}{5}
Triatomic linear7R2\frac{7R}{2}9R2\frac{9R}{2}97\frac{9}{7}
Triatomic non-linear3R3R4R4R43\frac{4}{3}

Spontaneity Crossover

Tcrossover=ΔHΔS(when both same sign)T_{crossover} = \frac{\Delta H}{\Delta S} \quad \text{(when both same sign)}

Useful Constants

R=8.314 J/(mol⋅K)=0.0821 L⋅atm/(mol⋅K)R = 8.314\ \text{J/(mol·K)} = 0.0821\ \text{L·atm/(mol·K)} 1 L⋅atm=101.3 J,ln2=0.693,ln10=2.3031\ \text{L·atm} = 101.3\ \text{J}, \quad \ln 2 = 0.693, \quad \ln 10 = 2.303

Like these notes? Save your own copy and start studying with NoteTube's AI tools.

Sign up free to clone these notes