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 : $\ln\,2 = 0693)$
- The rate constant, $k$, of the reaction is $13.86 \times 10^{-4} s ^{-1}$.
- Half-life of $X$ is $50\, s$.
- At $50\, s,-\frac{d[X]}{d t}=13.86 \times 10^{-3} mol\, L^{-1} s^{-1}$.
- At $100 \,s,-\frac{d[Y]}{d t}=3.46 \times 10^{-3} mol\, L^{-1} s^{-1}$.
The Correct Option is D
Solution and Explanation
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.
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Concepts Used:
Chemical Kinetics
Chemical kinetics is the description of the rate of a chemical reaction. This is the rate at which the reactants are transformed into products. This may take place by abiotic or by biological systems, such as microbial metabolism.
Rate of a Chemical Reaction:
The speed of a reaction or the rate of a reaction can be defined as the change in concentration of a reactant or product in unit time. To be more specific, it can be expressed in terms of: (i) the rate of decrease in the concentration of any one of the reactants, or (ii) the rate of increase in concentration of any one of the products. Consider a hypothetical reaction, assuming that the volume of the system remains constant. R → P
Read More: Chemical Kinetics MCQ
Factors Affecting The Reaction Rate:
- The concentration of Reactants - According to collision theory, which is discussed later, reactant molecules collide with each other to form products.
- Nature of the Reactants - The reaction rate also depends on the types of substances that are reacting.
- Physical State of Reactants - The physical state of a reactant whether it is solid, liquid, or gas can greatly affect the rate of change.
- Surface Area of Reactants - When two or more reactants are in the same phase of fluid, their particles collide more often than when either or both are in the solid phase or when they are in a heterogeneous mixture. In a heterogeneous medium, the collision between the particles occurs at an interface between phases. Compared to the homogeneous case, the number of collisions between reactants per unit time is significantly reduced, and so is the reaction rate.
- Temperature - If the temperature is increased, the number of collisions between reactant molecules per second. Increases, thereby increasing the rate of the reaction.
- Effect Of Solvent - The nature of the solvent also depends on the reaction rate of the solute particles.
- Catalyst - Catalysts alter the rate of the reaction by changing the reaction mechanism.