Question

Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. The rate-law expression is rate = k[CH3CHO]2, and k = 7.6 x 10-3 L/(mol.hr) at 527°C. What is the half-life of CH3CHO if 0.26 mole is injected into a 8.1 L vessel at 527°C?

Answer 1

The half-life of acetaldehyde (CH3CHO) undergoing decomposition in a second-order reaction at 527°C can be calculated using the half-life formula for a second-order reaction, t1/2 = 1 / (k*[A0]), by substituting the provided rate constant and initial concentration.

Step-by-step explanation:

The question regards determination of the half-life of acetaldehyde, CH3CHO, decomposing to methane and carbon monoxide in a second-order reaction. Given the reaction rate-law expression (rate = k[CH3CHO]2) and the rate constant (k = 7.6 x 10-3 L/(mol.hr)) at 527°C, we can calculate the half-life using the formula for the half-life of a second-order reaction, which is t1/2 = 1 / (k*[A0]), where [A0] is the initial concentration of the reactant.

The initial concentration is calculated by using the moles of CH3CHO and the volume of the container which are 0.26 moles and 8.1 L, respectively, yielding [A0] = 0.26 moles / 8.1 L. Substituting the given values into the half-life formula, we can find the half-life of CH3CHO.