Exams › JEE Advanced › Chemistry

For the first-order decomposition reaction N2O5 -> 2NO2 + (1/2)O2, the half-life is independent of pressure (confirming first-order kinetics). If the initial partial pressure of N2O5 is P0, which graph correctly represents the partial pressure of NO2 versus time?

1. (A) A curve for P_NO2 vs time that starts at P0 and decreases to zero
2. (B) A linearly increasing curve for P_NO2 vs time starting from P0
3. (C) A curve for P_NO2 vs time that increases asymptotically toward 2*P0
4. (D) An S-shaped curve for P_NO2 vs time that increases asymptotically toward 2*P0

Correct answer: (C) A curve for P_NO2 vs time that increases asymptotically toward 2*P0

Solution

Since reaction is first-order (constant half-life), P(N2O5) = P0*exp(-kt). P(NO2) = 2(P0 - P0*exp(-kt)) = 2*P0*(1-exp(-kt)), an exponential approach to 2*P0 (option C).

Related JEE Advanced Chemistry questions

• The given equation implies that:
• Given that C₁ represents the starting concentration of A, and C₁, C₂, and C₃ denote the concentrations of A, B, and C respectively at a specific time t, which equation can be used to calculate the values of k₁ and k₂?
• In the reaction M → N, the rate at which M is consumed becomes 8 times faster when the concentration of M is doubled. What is the reaction order with respect to M?
• For the reaction A(g) + B(g) ⇌ AB(g), the activation energy for the reverse reaction is greater than that for the forward reaction by 2RT (in J mol⁻¹). If the forward reaction's pre-exponential factor is four times that of the backward reaction, what is the magnitude of ΔG⁰ (in J mol⁻¹) at 300 K?
• A gaseous compound X decomposes into gaseous products Y and Z according to first-order kinetics (X -> Y + Z). Initially only X is present. Ten minutes after the reaction begins, the partial pressure of X is 200 mmHg and the total pressure of the mixture is 300 mmHg. What is the rate constant for this reaction?
• A catalyst is added to a first-order reaction at 500 K, causing its rate to become 2.718 times the original rate. The activation energy with the catalyst present is 4.15 kJ/mol. What was the activation energy before the catalyst was introduced? (Given: R = 8.3 J/K/mol, ln(2.718) = 1)

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →