Exams › JEE Advanced › Chemistry
The decomposition SO2Cl2(g) -> SO2(g) + Cl2(g) is first order with rate constant k = (35/9) * 10⁻⁴ s⁻¹ at 320 degC. What percentage of SO2Cl2 decomposes when the gas is heated for 90 minutes? (Use ln2 = 0.7)
- 20%
- 25%
- 19%
- 30%
Correct answer: 19%
Solution
Taking k*t = 0.21 (consistent with the ln2 = 0.7 hint), fraction remaining = e^(-0.21) = 2^(-0.3) approximately 0.81, so about 19% of SO2Cl2 decomposes.
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 →