A Rate constant decreases exponentially
B Rate constant increases exponentially
C Rate constant remains the same
D Rate constant increases linearly
Explanation: As T increases, Ea/RT decreases, making e^(-Ea/RT) larger, so k increases exponentially with temperature.
A Enthalpy when 1 mole burns in excess oxygen at standard conditions
B Enthalpy when all moles burn completely
C Enthalpy of formation of combustion products
D Enthalpy per gram of fuel burned
Explanation: Standard enthalpy of combustion is the enthalpy change when 1 mole of a substance undergoes complete combustion in excess O2 under standard conditions (298 K, 1 atm).
A Polyethylene
B Polypropylene
C Polytetrafluoroethylene (PTFE/Teflon)
D PVC
Explanation: Tetrafluoroethylene (CF2=CF2) undergoes addition polymerization to form PTFE (Teflon), where the double bond opens to form long chains.
A All electrons are paired
B Zero pairing energy because all t2g and eg orbitals are singly occupied
C Maximum splitting
D Delta_o equals the pairing energy
Explanation: In high-spin d5, each of the five d-orbitals gets one electron (t2g3 eg2) with no pairing, so no pairing energy is required.
A Nucleophilic attack on the carbocation
B Loss of the leaving group to form the carbocation
C Deprotonation
D Formation of the transition state
Explanation: The slowest step in SN1 is ionization: the departure of the leaving group to form a carbocation intermediate. This step has the highest activation energy.
A t1/2 = 0.693/k
B t1/2 = [A]0/2k
C t1/2 = 1/(k[A]0)
D t1/2 = 2.303/k
Explanation: For first-order reactions, t1/2 = ln(2)/k = 0.693/k, which is independent of the initial concentration.
A 3-Bromo-1-butene
B 2-Bromo-3-butene
C 1-Bromo-2-butene
D 3-Bromo-2-butene
Explanation: Numbering gives the double bond the lowest locant: C1=C2-C3(Br)-C4. Bromine is on C3, so the name is 3-bromobut-1-ene.
A n -> pi*
B pi -> pi*
C sigma -> sigma*
D n -> sigma*
Explanation: Sigma bonds are the strongest, so sigma -> sigma* transitions require the highest energy (shortest wavelength), typically in the far UV region.
A Less than 7
B Equal to 7
C Greater than 7
D Depends on the indicator
Explanation: The salt formed (e.g., NaCl) is neutral, so the pH at equivalence is exactly 7 for strong acid-strong base titrations.
A A ligand that forms one bond to the metal
B A ligand that forms two bonds to the metal
C A ligand that forms six bonds to the metal
D A ligand that carries a -2 charge
Explanation: A bidentate ligand has two donor atoms that simultaneously coordinate to the metal center, forming a chelate ring (e.g., ethylenediamine, oxalate).
A lambda = h/mv
B lambda = h x mv
C lambda = mv/h
D lambda = h/2mv
Explanation: The De Broglie equation states lambda = h/p = h/(mv), where h is Planck\'s constant, m is mass, and v is velocity, predicting wave-particle duality.
A Bromination of benzene
B Addition of HBr to propene
C Chlorination of methane
D Hydrolysis of an ester
Explanation: HBr adds across the C=C double bond in propene. The electrophilic H+ attacks the pi bond first, followed by nucleophilic Br- attack.
A m = (M x I x t) / (n x F)
B m = (n x F) / (M x I x t)
C m = (M x F) / (n x I x t)
D m = (I x t) / (M x n x F)
Explanation: Faraday\'s first law: mass deposited (m) = (M x I x t)/(n x F), where M is molar mass, I is current, t is time, n is electrons transferred, and F = 96485 C/mol.
A IR spectroscopy only
B 1H NMR spectroscopy
C UV-Vis spectroscopy
D Flame test
Explanation: 1H NMR gives different splitting patterns: 1-propanol shows a triplet for the CH3, while 2-propanol shows a distinctive septet for the CH and a doublet for two equivalent CH3 groups.
A Unimolecular
B Bimolecular
C Termolecular
D Quadrimolecular
Explanation: Molecularity is the number of molecules that come together in an elementary step. Three simultaneous collisions give termolecular (trimolecular) molecularity.
A Kp = Kc
B Kp = Kc(RT)^2
C Kp = Kc(RT)^(-2)
D Kp = Kc(RT)^4
Explanation: Kp = Kc(RT)^(Delta n). Delta n = 2 - (1+3) = -2, so Kp = Kc(RT)^(-2) = Kc/(RT)^2.
A Tertiary > Secondary > Primary
B Primary > Secondary > Tertiary
C Secondary > Primary > Tertiary
D All equal
Explanation: E2 reactivity increases with alkyl substitution because more substituted alkenes (Zaitsev product) are more stable, and tertiary substrates have more beta-hydrogens.
A mu = sqrt(n(n+2)) BM
B mu = n(n+2) BM
C mu = sqrt(n+2) BM
D mu = 2sqrt(n) BM
Explanation: The spin-only formula gives magnetic moment mu = sqrt(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.
A Cyclohexanol
B Adipic acid (hexanedioic acid)
C Cyclohexanone
D No reaction
Explanation: Hot, concentrated KMnO4 is a strong oxidizing agent that cleaves the C=C double bond in cyclohexene, oxidizing it completely to adipic acid (HOOC-CH2-CH2-CH2-CH2-COOH).
A Planar (trigonal)
B Trigonal bipyramidal (pentacoordinate carbon)
C Tetrahedral
D Linear
Explanation: The SN2 transition state has a pentacoordinate carbon with the nucleophile and leaving group on opposite sides, forming a trigonal bipyramidal arrangement.