All Free Chemistry MCQs with Answers

181. The mathematical statement of the first law of thermodynamics is

• A. change in internal energy equals heat added plus work done on the system
• B. change in internal energy equals heat added minus temperature
• C. enthalpy equals internal energy divided by pressure
• D. heat equals work in every process

Explanation: The internal energy of a system can be raised either by heating it or by doing work on it, and the equation simply accounts for both routes. When the system does work on the surroundings, for example by expanding against the atmosphere, that term becomes negative. The sign conventions must be stated clearly, since they are the commonest source of error.

182. Which of the following is NOT a state function?

• A. Enthalpy
• B. Internal energy
• C. Work
• D. Temperature

Explanation: The amount of work done in going from one state to another depends on the route taken, for example whether a gas expands quickly or slowly, so work is a path function. Enthalpy, internal energy and temperature depend only on the current condition of the system. Heat is the other common path function, which is why heat and work are written as small quantities rather than as changes in a property.

183. The dissolving of ammonium nitrate in water lowers the temperature of the solution, which means that

• A. the process is exothermic
• B. no energy change occurs
• C. the lattice energy is zero
• D. the energy absorbed in breaking the lattice exceeds the energy released on hydrating the ions

Explanation: Dissolving involves two steps with opposite signs: separating the ions from the lattice costs energy while surrounding them with water molecules releases it, and the overall sign depends on which is larger. For ammonium nitrate the lattice term wins, so the process is endothermic and the solution cools, which is what an instant cold pack exploits. For calcium chloride the balance goes the other way and the solution warms.

184. An energy profile diagram for an endothermic reaction shows

• A. products at a higher energy level than reactants, with an activation barrier in between
• B. products at a lower energy level than reactants
• C. no activation energy
• D. reactants and products at the same level

Explanation: In an endothermic reaction energy is absorbed overall, so the products lie above the reactants and the enthalpy change is positive, but the curve must still rise to a peak because the activation barrier has to be crossed. The barrier for the reverse, exothermic direction is correspondingly smaller. Reading the enthalpy change as the difference between the two flat levels, not as the height of the peak, is the key skill.

185. The carbon atom bonded to the halogen in an alkyl halide is attacked by nucleophiles because it carries

• A. a partial positive charge, since the halogen is more electronegative
• B. a partial negative charge
• C. a full negative charge
• D. no charge at all

Explanation: The carbon to halogen bond is polar, with electron density drawn towards the halogen, so the carbon is electron deficient and attracts species with a lone pair. This polarity is the reason alkyl halides are far more reactive than alkanes despite looking similar. The halogen leaves as a stable halide ion, which is why it is described as a good leaving group.

186. The compound (CH3)3CBr is classified as

• A. a primary alkyl halide
• B. a secondary alkyl halide
• C. a tertiary alkyl halide
• D. an aryl halide

Explanation: The carbon carrying the bromine is attached to three other carbon atoms, which makes it tertiary, whereas a primary carbon is attached to one and a secondary to two. This classification matters because tertiary halides react overwhelmingly by the SN1 route while primary halides prefer SN2. An aryl halide would have the halogen attached directly to a benzene ring.

187. The order of reactivity of alkyl halides towards nucleophilic substitution, for the same alkyl group, is

• A. R-F greater than R-Cl greater than R-Br greater than R-I
• B. R-I greater than R-Br greater than R-Cl greater than R-F
• C. all four react at the same rate
• D. R-Cl greater than R-I greater than R-Br greater than R-F

Explanation: Reactivity is governed by the strength of the carbon to halogen bond, and that bond weakens down the group as the halogen gets larger, so the iodide breaks most easily and the fluoride hardly reacts at all. This outweighs the fact that fluorine creates the largest dipole. Iodide is also the most stable leaving group of the four.

188. The reaction of bromoethane with aqueous potassium hydroxide gives

• A. ethene
• B. ethanoic acid
• C. ethane
• D. ethanol

Explanation: In aqueous solution the hydroxide ion acts as a nucleophile, replacing the bromine to give ethanol in a substitution reaction. The same reagent in hot ethanolic solution behaves as a base instead, removing a hydrogen and giving ethene by elimination, so the solvent decides the outcome. This pair of reactions is examined constantly and turns entirely on the conditions.

189. An SN2 reaction is characterised by

• A. a two step mechanism through a carbocation
• B. first order kinetics involving only the alkyl halide
• C. a single step in which the nucleophile attacks as the leaving group departs, with second order kinetics
• D. the formation of a free radical intermediate

Explanation: The nucleophile approaches from the side opposite the leaving group and bond forming and bond breaking occur together through a five coordinate transition state, so the rate depends on the concentrations of both the halide and the nucleophile. Because attack is from the back, the configuration at that carbon is inverted, an effect known as Walden inversion. Primary halides follow this route because they are least hindered.

190. A tertiary alkyl halide usually undergoes nucleophilic substitution by the SN1 mechanism because

• A. the tertiary carbocation formed is relatively stable and the carbon is too hindered for back side attack
• B. tertiary halides have the weakest carbon to halogen bond
• C. tertiary carbon atoms carry a full positive charge already
• D. the nucleophile is always weak

Explanation: Three electron releasing alkyl groups spread the positive charge and stabilise the carbocation, so the slow ionisation step becomes feasible, and at the same time those bulky groups block the approach a concerted SN2 attack would need. The rate therefore depends only on the halide concentration, giving first order kinetics. Because the planar carbocation can be attacked from either face, a racemic mixture results.