Exams › JEE Advanced › Chemistry
Milk is pasteurised when heated at 67 deg C for 4 hours. If Ea/2.303 for the process equals 23.8 kcal/mol, what is the minimum time (in seconds) required for pasteurisation at 77 deg C? [R = 2 cal/K-mol]
- 1440
- 14400
- 720
- 144
Correct answer: 1440
Solution
The rate constant increases tenfold from 340 K to 350 K, so the time falls to one-tenth of 4 hours, i.e. 0.4 h = 1440 s.
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 →