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Chem test 3
| Question | Answer |
|---|---|
| Change in entropy equation | ΔS=ΔSf-ΔSi/ /ΔS=qrev/T (constant T) |
| Change in entropy of surrounding equation | ΔSsurr=(-ΔHsys)/T (constant P and T) |
| Enthalpy change | ΔH |
| Entropy change | ΔS |
| Entropy equation | S=(k)ln(W) |
| Work equation | wf/wi |
| qrev equation | qrev=nxΔHvap |
| Change in entropy of universe equation | ΔSuni=ΔSsys+ΔSsurro (must be +) |
| qsurr=-qsys | - |
| Boltzmann contant | k=1.38x10^-23 |
| Change in entropy of system equation | ΔS*rxn= ∑[np(S*{products}]- ∑[nr(S*{reactants}] |
| Change in Gibbs free energy equation | ΔG=ΔHsys-T(ΔSsys) |
| Gibbs free energy spontaneous | ΔG<0 |
| Gibbs free energy nonspontaneous | ΔG>0 |
| Gibbs free energy equilibrium | ΔG=0 |
| Change in Gibbs free energy of reaction (equation 2) (@25 *C) | ΔG*rxn= ∑[n(ΔG*f{products}]- ∑[n(ΔG*f{reactants}] |
| Total change in Gibbs free energy | ΔG*tot.=ΔG*react1+ΔG*react2+... |
| Equilibrium constant equation (K) | aA+bB <-> cC+dD (only aqueous solutions and gases) K= [C]^c[D]^d/[A]^a[B]^b |
| Reverse reaction is favored. More reactants than products | K<<1 |
| Forward reaction is favored. More products than reactants | K>>1 |
| Nether direction is favored. | K=1 |
| Reaction quotient (Qc) | Using K equation with the initial concentrations |
| System goes from left to right to reach equil. | Qc<Kc |
| System goes from right to left to reach equil. | Qc>Kc |
| System @ equil. | Qc=Kc |
| ΔG under nonstandard conditions | ΔG=ΔG*+(R)(T)lnQ T= in K |
| R value for J/K | R=8.314 J/K |
| ΔHrxn equation | ΔHrxn=∑products(npΔHf)−∑reactants(nrΔHf) |
| When ΔH is negative and ΔS is positive | ΔG is negative at all temperatures |
| When ΔH is positive and ΔS is negative | ΔG is positive at all temperatures. |
| When both ΔH and ΔS are negative | ΔG will be negative at low temperatures and positive at high temperatures |
| When both ΔH and ΔS are positive | ΔG will be negative at high temperatures and positive at low temperatures. |
Created by:
mandi239
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