Which of the following graphical relations correctly characterizes a first-order reaction?

A plot of log(a - x) versus t is a straight line with a negative slope
A plot of log(a - x) versus 1/t is a straight line with a positive slope
A plot of t(1/2) versus a is a horizontal line
Both option 1 and option 3

Correct answer: Both option 1 and option 3

Solution

For a first-order reaction, log(a-x) versus t is linear with negative slope -k/2.303, and t(1/2) = 0.693/k is independent of a, giving a horizontal line. Both statements 1 and 3 are correct.

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 →