chemistry : class12 Gas evolution rate : Preparation of phosphine : p-Block Elements MCQ Questions and Answers
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Gas evolution rate
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MCQ Questions
1. In the preparation of phosphine by heating white phosphorus with concentrated NaOH solution, which change generally increases the initial rate of phosphine evolution without changing the overall stoichiometric equation?
A.
Increasing the surface area of white phosphorus
B.
Replacing water with an equal volume of kerosene
C.
Adding sodium hypophosphite while keeping all other conditions constant
D.
Cooling the reaction mixture substantially
Correct Answer:
Increasing the surface area of white phosphorus
Explanation:
A larger surface area provides more contact between phosphorus and the alkaline solution, increasing the rate of the heterogeneous reaction and hence the initial rate of PH3 evolution. The stoichiometric equation remains unchanged.
2. For the reaction P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2, the volume of phosphine collected is 90 mL after 6.0 min and 150 mL after 10.0 min. What is the average phosphine evolution rate between 6.0 and 10.0 min?
A.
6.0 mL min−1
B.
9.0 mL min−1
C.
15.0 mL min−1
D.
24.0 mL min−1
Correct Answer:
15.0 mL min−1
Explanation:
The average rate is calculated from the increase in gas volume divided by the time interval: (150 − 90) mL ÷ (10.0 − 6.0) min = 60 ÷ 4.0 = 15.0 mL min−1.
3. Two identical phosphine generators are operated with equal masses of white phosphorus. Generator I contains 2.0 mol L−1 NaOH at 60 °C, whereas Generator II contains 2.0 mol L−1 NaOH at 25 °C. Which observation is most appropriate?
A.
Generator I evolves phosphine more rapidly initially, but the final amount is determined by the limiting reactant
B.
Generator II evolves phosphine more rapidly because lower temperature prevents PH3 decomposition
C.
Both generators evolve phosphine at the same rate because temperature does not affect reaction rate
D.
Generator I produces twice the final amount of phosphine solely because its temperature is higher
Correct Answer:
Generator I evolves phosphine more rapidly initially, but the final amount is determined by the limiting reactant
Explanation:
Increasing temperature generally increases the reaction rate, so the initial gas evolution is faster in Generator I. However, the final quantity of PH3 depends on the amounts of reactants and stoichiometry, not simply on the temperature.
4. A phosphine-preparation experiment gives the following gas volumes at equal 2-minute intervals: 0, 44, 76, 98, 112, and 120 mL. Which interval has the greatest average gas evolution rate?
A.
0–2 min
B.
2–4 min
C.
4–6 min
D.
8–10 min
Correct Answer:
0–2 min
Explanation:
The volume increases are 44, 32, 22, 14, and 8 mL over equal time intervals. Therefore, the greatest average rate occurs during the first 2 minutes, when the increase is 44 mL.
5. During phosphine generation from white phosphorus and aqueous NaOH, the gas-evolution curve gradually becomes less steep even though unreacted phosphorus remains visible. Which explanation best accounts for the decreasing rate?
A.
The effective contact between phosphorus and the alkaline solution decreases as the reaction proceeds
B.
The stoichiometric coefficient of phosphine decreases continuously during the reaction
C.
Phosphine changes into sodium hypophosphite without consuming phosphorus
D.
The molar volume of every gas necessarily decreases with time at constant temperature
Correct Answer:
The effective contact between phosphorus and the alkaline solution decreases as the reaction proceeds
Explanation:
Gas evolution can slow when the accessible reactive surface or contact between the solid phosphorus and solution decreases, for example because of aggregation or altered wetting. This affects the rate, while the stoichiometric equation remains fixed.
6. In two trials for phosphine preparation, all conditions are identical except that trial X uses phosphorus pieces of 1 mm average size and trial Y uses pieces of 5 mm average size. If the same total mass reacts completely, which result is expected?
A.
X has a faster initial PH3 evolution and reaches the same final gas volume as Y
B.
Y has a faster initial PH3 evolution and reaches a larger final gas volume than X
C.
X and Y have the same initial rate but X gives twice the final gas volume
D.
X gives no phosphine because smaller particles cannot undergo disproportionation
Correct Answer:
X has a faster initial PH3 evolution and reaches the same final gas volume as Y
Explanation:
Smaller phosphorus pieces provide greater total surface area, increasing the initial heterogeneous reaction rate. Since the reacting mass and composition are the same, complete reaction produces the same total amount of PH3.
7. At a particular stage of phosphine preparation, the instantaneous gas-evolution rate is 12 mL min−1. Assuming the rate remains constant for the next 5 minutes, what volume of phosphine is expected to be collected during that period?
A.
2.4 mL
B.
17 mL
C.
60 mL
D.
144 mL
Correct Answer:
60 mL
Explanation:
For a constant rate, gas volume evolved equals rate multiplied by time: 12 mL min−1 × 5 min = 60 mL. This is a rate-to-volume calculation for PH3 evolution.
8. A student doubles the concentration of NaOH in a phosphine generator while keeping the mass and surface area of phosphorus, temperature, and total liquid volume constant. Which statement is scientifically safest regarding the initial rate?
A.
The initial rate may increase because hydroxide participates in the reaction, but the exact factor cannot be inferred from stoichiometry alone
B.
The initial rate must exactly double because the coefficient of NaOH is three
C.
The initial rate must become one-half because the reaction consumes water
D.
The initial rate cannot change because NaOH is not present in phosphine
Correct Answer:
The initial rate may increase because hydroxide participates in the reaction, but the exact factor cannot be inferred from stoichiometry alone
Explanation:
Hydroxide is a reactant in phosphine preparation, so changing its concentration can affect the reaction rate. However, stoichiometric coefficients do not by themselves give reaction orders, so an exact rate change cannot be concluded.
9. A gas-collection setup for phosphine preparation has a constant leak of 2.0 mL min−1. The measured volume increases by 18 mL in 3.0 min. What was the approximate actual PH3 evolution rate during this interval?
A.
4.0 mL min−1
B.
6.0 mL min−1
C.
8.0 mL min−1
D.
20.0 mL min−1
Correct Answer:
8.0 mL min−1
Explanation:
The measured net rate is 18 ÷ 3.0 = 6.0 mL min−1. Since 2.0 mL min−1 is lost through leakage, the actual evolution rate is 6.0 + 2.0 = 8.0 mL min−1.
10. For two runs of the same phosphine-preparation reaction, the tangent to the volume–time curve at 4 minutes has slopes of 7.5 mL min−1 in run I and 3.0 mL min−1 in run II. Which conclusion is correct?
A.
Run I has the greater instantaneous phosphine evolution rate at 4 minutes
B.
Run II has produced a greater total amount of phosphine by 4 minutes
C.
The final phosphine volume in run I must be 2.5 times that in run II
D.
The stoichiometric equation is different in the two runs
Correct Answer:
Run I has the greater instantaneous phosphine evolution rate at 4 minutes
Explanation:
The slope of a volume–time curve at a particular time represents the instantaneous gas-evolution rate. Since 7.5 mL min−1 is greater than 3.0 mL min−1, run I evolves PH3 faster at 4 minutes; this does not by itself determine the total or final volume.
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