A substance dissociates through a first-order reaction with a half-life of 32 ext min. If the initial concentration of the substance is 2.7 imes 1010 ext molecules/L, what will be the concentration 2.

Here are the solutions for the questions in the image. 2ai) A substance dissociates through a first-order reaction with a half-life of 32 min. If the initial concentration of the substance is 2.7 × 10^10 molecules/L, what will be the concentration 2.24 h later? Step 1: Convert the given time to minutes. The half-life is in minutes, so we convert the total time from hours to minutes for consistency. t = 2.24 h × 60 minh = 134.4 min Step 2: Calculate the rate constant (k) for the first-order reaction. For a first-order reaction, the half-life (t_1/2) is related to the rate constant (k) by the formula: t_1/2 = ((2))/(k) Rearranging to solve for k: k = ((2))/(t_1/2) k = (0.693)/(32 min) k ≈ 0.02166 min^-1 Step 3: Use the integrated rate law for a first-order reaction to find the concentration at time t. The integrated rate law is: ([A]_t) = ([A]_0) - kt Where [A]_t is the concentration at time t, and [A]_0 is the initial concentration. This can also be written as: [A]_t = [A]_0 e^-kt Step 4: Substitute the known values into the integrated rate law. [A]_t = (2.7 × 10^10 molecules/L) × e^-(0.02166 min^-1 × 134.4 min) [A]_t = (2.7 × 10^10 molecules/L) × e^-2.9076 [A]_t = (2.7 × 10^10 molecules/L) × 0.05462 [A]_t = 1.47474 × 10^9 molecules/L Step 5: Round the final answer to two significant figures, consistent with the given data. [A]_t = 1.5 × 10^9 molecules/L ii) Express the rate of the following reaction in terms of the formation of ammonia: N_2(g) + 3H_2(g) 2NH_3(g). Step 1: Write the general rate expression for the reaction. The rate of reaction is expressed as the change in concentration of a reactant or product over time, divided by its stoichiometric coefficient. Reactants have a negative sign because their concentration decreases. Rate = -d[N_2]dt = -(1)/(3)d[H_2]dt = (1)/(2)d[NH_3]dt Step 2: Express the rate specifically in terms of the formation of ammonia. The formation of ammonia is represented by the term d[NH_3]dt. To express the overall rate of reaction in terms of this, we use the coefficient. Rate = (1)/(2)d[NH_3]dt 3 done, 2 left today. You're making progress.