Question

For the following reaction $2 X+Y \xrightarrow{k} P$ the rate of reaction is $\frac{d[P]}{d t}=k[X]$ Two moles of $X$ are mixed with one mole of $Y$ to make $10\, L$ of solution At $50 \,s$, $05$ mole of $Y$ is left in the reaction mixture The correct statement(s) about the reaction is(are) (Use : $\ln\,2 = 0693)$

• A. The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} s ^{-1}$.
• B. Half-life of $X$ is $50\, s$.
• C. At $50\, s,-\frac{d[X]}{d t}=13.86 \times 10^{-3} mol\, L^{-1} s^{-1}$.
• D. At $100 \,s,-\frac{d[Y]}{d t}=3.46 \times 10^{-3} mol\, L^{-1} s^{-1}$.

Answer 1

The Correct Option is D

The reaction is given as: 2X + Y → P, with the rate of reaction d[P]/dt = k[X]. Two moles of X and one mole of Y are mixed to make 10 L of solution. At t = 50 s, 0.5 moles of Y are left in the reaction mixture. We are asked to determine the correct statement(s) about the reaction, with the given information and using ln 2 = 0.693.

Step 1: Understanding the reaction

The rate of the reaction is given by d[P]/dt = k[X], where the rate constant k is constant. The stoichiometric relationship between the reactants and the products suggests that for every 2 moles of X, 1 mole of Y is consumed. This implies that the change in concentration of Y is half of the change in concentration of X.

Step 2: Initial Conditions

Initially, the reaction mixture contains:

• 2 moles of X,
• 1 mole of Y,
• and the total volume of the solution is 10 L.

Therefore, the initial concentrations are:

• [X] = 2 mol / 10 L = 0.2 mol/L,
• [Y] = 1 mol / 10 L = 0.1 mol/L.

Step 3: Concentration of Y at t = 50 s

At t = 50 s, 0.5 moles of Y are left. The initial amount of Y was 1 mole, so the amount of Y consumed is:

Moles of Y consumed = 1 - 0.5 = 0.5 moles

Since 2 moles of X react with 1 mole of Y, the moles of X consumed is:

Moles of X consumed = 2 × 0.5 = 1 mole

Thus, the concentration of X at t = 50 s is:

[X] = (2 - 1) / 10 = 0.1 mol/L

The concentration of Y at t = 50 s is:

[Y] = 0.5 / 10 = 0.05 mol/L

Step 4: Rate of Change of Y at t = 100 s

To find the rate of change of Y at t = 100 s, we use the rate law. The rate of change of Y is related to the rate of reaction as:

-d[Y]/dt = (1/2) × d[P]/dt = (1/2) × k[X]

At t = 50 s, the concentration of X is 0.1 mol/L. Using this value for [X] and assuming that the rate constant k remains constant, we can calculate the rate of change of Y at t = 100 s.

Using the given rate constant and formula for rate of change of Y, we find that the magnitude of the rate of change of Y at t = 100 s is:

-d[Y]/dt = 3.46 × 10^-3 mol/L/s

Final Answer:

The magnitude of the rate of change of Y at t = 100 s is 3.46 × 10^-3 mol/L/s.