Exams › JEE Advanced › Chemistry

In the reaction xA -> yB, it is given that log10(-d[A]/dt) = log10(d[B]/dt) + 0.3, where the negative sign indicates the rate of disappearance of reactant. Find x: y. (Use log10(2) = 0.3.)

1. 1: 2
2. 2: 1
3. 3: 1
4. 3: 10

Correct answer: 2: 1

Solution

From the given equation, -d[A]/dt = 2 * d[B]/dt. Stoichiometry requires -d[A]/dt: d[B]/dt = x: y = 2: 1.

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 →