All Free Chemistry MCQs with Answers

21. The average kinetic energy of the molecules of an ideal gas depends only on

• A. pressure
• B. absolute temperature
• C. volume
• D. molar mass

Explanation: Average kinetic energy equals three halves of kT, so it is fixed by absolute temperature alone. This is why hydrogen and oxygen at the same temperature have the same average kinetic energy even though hydrogen molecules move much faster, being lighter. Molar mass affects speed, not energy.

22. Real gases deviate most from ideal behaviour at

• A. high temperature and low pressure
• B. low temperature and high pressure
• C. high temperature and high pressure
• D. standard temperature and pressure

Explanation: At high pressure the molecules are close enough that their own volume is no longer negligible, and at low temperature they move slowly enough for intermolecular attractions to matter. Both assumptions of the ideal model fail together under those conditions, which is also why gases liquefy there. A gas behaves most ideally when it is hot and dilute.

23. In the van der Waals equation, the constant a corrects for

• A. the volume occupied by the gas molecules
• B. the intermolecular forces of attraction
• C. the number of moles present
• D. the temperature of the gas

Explanation: The term involving a raises the measured pressure to what it would have been without attractions pulling molecules away from the wall, so a is larger for gases that are easily liquefied. The constant b is the excluded volume correction for the finite size of the molecules. Moles and temperature already appear explicitly in the equation.

24. What volume is occupied by 0.5 mole of an ideal gas at STP?

• A. 5.6 dm3
• B. 11.2 dm3
• C. 22.4 dm3
• D. 44.8 dm3

Explanation: One mole of any ideal gas occupies 22.4 dm3 at STP, so half a mole occupies 11.2 dm3. The identity of the gas is irrelevant, which is the point of Avogadro's law. The value 5.6 dm3 would correspond to a quarter of a mole.

25. A chemical system is said to be in dynamic equilibrium when

• A. both reactions have stopped completely
• B. the forward and reverse reactions continue at equal rates
• C. the amounts of reactants and products are equal
• D. all the reactants have been converted to products

Explanation: The word dynamic means both reactions are still running; they simply cancel out, so the concentrations stay constant while molecules continue to convert in both directions. Equal concentrations of reactant and product are not required and are usually not the case, which is the standard misconception. Equilibrium can only be reached in a closed system.

26. For the reaction N2 + 3H2 gives 2NH3, the equilibrium constant expression Kc is

• A. [N2][H2]^3 divided by [NH3]^2
• B. [NH3] divided by [N2][H2]
• C. [NH3]^2 divided by [N2][H2]^3
• D. 2[NH3] divided by [N2] + 3[H2]

Explanation: Kc is written as the product concentrations over the reactant concentrations, each raised to the power of its coefficient in the balanced equation. Inverting the expression gives the equilibrium constant for the reverse reaction, which is why the first option is offered. Coefficients become exponents, never multipliers, which rules out the last option.

27. A very large value of the equilibrium constant Kc indicates that

• A. the reaction reaches equilibrium very quickly
• B. the reaction is exothermic
• C. the reactants are favoured at equilibrium
• D. the products are strongly favoured at equilibrium

Explanation: Kc compares product to reactant concentrations, so a large value means the equilibrium position lies far to the right and the reaction goes almost to completion. Kc says nothing at all about how fast equilibrium is reached, which is a matter of kinetics, and this is the confusion the question targets. The Haber process is the classic case of a favourable equilibrium that is nevertheless slow without a catalyst.

28. Le Chatelier's principle states that when a stress is applied to a system at equilibrium, the system

• A. shifts in the direction that relieves the stress
• B. shifts in the direction that increases the stress
• C. stops reacting altogether
• D. always shifts towards the products

Explanation: The position of equilibrium moves so as to partly oppose the change imposed, whether that change is in concentration, pressure or temperature. Adding a reactant therefore drives the reaction forward, and removing a product does the same. The principle predicts direction only; it says nothing about how far or how fast the shift occurs.

29. For the equilibrium N2 + 3H2 gives 2NH3, which is exothermic, the yield of ammonia is increased by

• A. raising the temperature and lowering the pressure
• B. raising the pressure and lowering the temperature
• C. adding a catalyst
• D. removing nitrogen from the mixture

Explanation: There are four moles of gas on the left and two on the right, so high pressure pushes the equilibrium towards the smaller volume, and since the forward reaction releases heat a lower temperature also favours it. A catalyst speeds the approach to equilibrium without changing the yield at all, which is the standard trap. In practice a compromise temperature near 450 degrees Celsius is used because the reaction is too slow when cold.

30. Increasing the pressure has NO effect on the position of the equilibrium H2 + I2 gives 2HI because

• A. the reaction is exothermic
• B. hydrogen iodide is a gas
• C. the number of moles of gas is the same on both sides
• D. the reaction has no catalyst

Explanation: Pressure changes shift an equilibrium only when the two sides differ in the number of gas molecules, and here there are two moles on each side, so neither direction relieves the stress. The rate of reaching equilibrium does rise with pressure, but the position stays put. Temperature changes will still shift this equilibrium, since the two directions differ in enthalpy.