Reactivity 2: How much, how fast and how far? practice questions.

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. (Ar: C = 12.0, H = 1.0, O = 16.0.) What is its empirical formula?

• CH2O
• C2H4O2
• CHO
• C2H6O

Answer: CH2O

Divide each percentage by its Ar: C = 40.0/12.0 = 3.33, H = 6.7/1.0 = 6.7, O = 53.3/16.0 = 3.33. Divide by the smallest (3.33): C = 1, H = 2, O = 1. The empirical formula is CH2O.

For N2 + 3H2 -> 2NH3, if 2.0 mol N2 reacts with 3.0 mol H2, which is the limiting reactant and how much NH3 forms?

• N2 limiting, 4.0 mol NH3
• H2 limiting, 2.0 mol NH3
• H2 limiting, 3.0 mol NH3
• N2 limiting, 2.0 mol NH3

Answer: H2 limiting, 2.0 mol NH3

The ratio needed is 1 N2 : 3 H2. 2.0 mol N2 would need 6.0 mol H2, but only 3.0 mol H2 is available, so H2 is limiting. 3.0 mol H2 x (2 NH3 / 3 H2) = 2.0 mol NH3.

At STP (0 degrees C, 100 kPa), the molar volume of an ideal gas is 22.7 dm3/mol. What volume does 0.50 mol of oxygen gas occupy at STP?

• 11.35 dm3
• 22.7 dm3
• 45.4 dm3
• 5.68 dm3

Answer: 11.35 dm3

At STP the volume of any ideal gas = moles x molar volume = 0.50 x 22.7 = 11.35 dm3. The identity of the gas does not matter for an ideal gas.

A fixed amount of gas at constant temperature has volume 2.0 dm3 at 100 kPa. What is its volume when the pressure is increased to 250 kPa?

• 0.80 dm3
• 1.25 dm3
• 5.0 dm3
• 0.50 dm3

Answer: 0.80 dm3

By Boyle's law, P1V1 = P2V2 at constant T. V2 = (P1V1)/P2 = (100 x 2.0)/250 = 200/250 = 0.80 dm3.

On a Maxwell-Boltzmann energy distribution, what happens when temperature is increased?

• The curve shifts towards lower energies and its peak becomes taller and much narrower
• The peak shifts to higher energy and more molecules exceed the activation energy
• The total area under the curve increases
• The activation energy decreases

Answer: The peak shifts to higher energy and more molecules exceed the activation energy

At higher temperature the distribution broadens and its peak shifts to higher energy (and becomes lower). A larger fraction of molecules has energy greater than or equal to the activation energy, so the reaction rate increases. The total area (number of molecules) stays constant.

For a reaction A + B -> products, doubling [A] doubles the rate and doubling [B] quadruples the rate. What is the overall order of reaction?

• 1
• 2
• 3
• 0

Answer: 3

Doubling [A] doubles the rate: first order in A. Doubling [B] quadruples the rate (2^2): second order in B. Rate = k[A][B]^2. Overall order = 1 + 2 = 3.