Question 1Level 2 · ExamSingle correct
The gas-phase decomposition 2N2O5(g)→4NO2(g)+O2(g) is carried out in a closed rigid vessel at constant temperature, starting with pure N2O5. At a certain instant the total pressure is increasing at 0.090 atm min−1. Assume ideal behaviour and that no other reaction occurs. At that instant, the rate of decrease of the partial pressure of N2O5 is ______. Exactly one option is correct.
A0.030 atm min−1
B0.060 atm min−1
C0.045 atm min−1
D0.180 atm min−1
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Answer: B
Define the rate in pressure units as r=−21dtdpN2O5=41dtdpNO2=dtdpO2.
At constant V and T, partial pressures add: dtdPtot=−2r+4r+r=3r.
So r=0.090/3=0.030 atm min−1 and −dtdpN2O5=2r=0.060 atm min−1 (option B).
Trap: 0.030 is the rate of reaction r, not the rate of loss of N2O5; 0.180=2×0.090 forgets to divide by the net pressure change of 3 per unit of reaction.
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
For a gas-phase reaction A(g)→ products of order 23 in A at constant volume and temperature T, the rate may be written either as −dtdpA=kppA3/2 (partial pressure pA) or as −dtd[A]=kc[A]3/2 (molar concentration). Treating A as an ideal gas (pA=[A]RT), kc is related to kp by ______. Exactly one option is correct.
Akc=kp(RT)1/2
Bkc=kp(RT)−1/2
Ckc=kp(RT)3/2
Dkc=kp
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Answer: A
With p=[A]RT at fixed T: −dtdp=RT(−dtd[A]).
So RT(−dtd[A])=kp([A]RT)3/2, giving −dtd[A]=kp(RT)1/2[A]3/2.
Hence kc=kp(RT)1/2 (option A). In general kc=kp(RT)n−1 for order n.
Unit check: kp has units atm−1/2 s−1; (RT)1/2 has units (atm L mol−1)1/2, so kc has units L1/2 mol−1/2 s−1, as expected for order 3/2.
Difficulty: Very hardQuestion 3Level 2 · ExamNumerical answer
For the reaction A+2B→4C at 300 K, the rate law has the form r=k[A]x[B]y. The following initial rates of formation of C, d[C]/dt, were measured:
Expt 1: [A]0=0.10 M, [B]0=0.10 M, d[C]/dt=2.0×10−4 mol L−1 s−1
Expt 2: [A]0=0.40 M, [B]0=0.10 M, d[C]/dt=4.0×10−4 mol L−1 s−1
Expt 3: [A]0=0.40 M, [B]0=0.30 M, d[C]/dt=3.6×10−3 mol L−1 s−1
In a fourth experiment at the same temperature with [A]0=0.025 M and [B]0=0.50 M, the initial rate of consumption of B, −d[B]/dt, in units of 10−4 mol L−1 s−1, is ______. Give the answer correct to two decimal places.
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Answer: 12.5
Expt 1 → 2: [A] ×4 at fixed [B], rate ×2, so 4x=2, x=21.
Expt 2 → 3: [B] ×3 at fixed [A], rate ×9, so y=2.
Expt 4 relative to Expt 1: dtd[C]=2.0×10−4×(0.25)1/2×(5)2=2.5×10−3 mol L−1 s−1.
From stoichiometry, −dtd[B]=42dtd[C]=1.25×10−3 mol L−1 s−1 =12.50×10−4.
Answer: 12.50.
Difficulty: Hard