X(g) + 2 Y(g) → XY_2(g) In order to determine the order of the reaction represented above, the initi — Kinetics Chemistry Question

STEPS:

1. Understand the definition of the rate of consumption (disappearance) of a reactant: In chemical kinetics, the rate of a reaction can be expressed in terms of the rate of change of concentration of any reactant or product, scaled by the reciprocal of its stoichiometric coefficient in the balanced chemical equation:
$$\text{Rate of reaction} = -\frac{\Delta[\text{X}]}{\Delta t} = -\frac{1}{2}\frac{\Delta[\text{Y}]}{\Delta t}$$
2. Examine the stoichiometry of the balanced chemical equation:
The reaction equation is:
$$\text{X}(g) + 2\ \text{Y}(g) \rightarrow \text{XY}_2(g)$$
* The coefficient of $\text{X}$ is 1.
* The coefficient of $\text{Y}$ is 2.
* This means that for every 1 mole of $\text{X}$ that is consumed, exactly 2 moles of $\text{Y}$ must be consumed in the same time interval.
3. Formulate the mathematical relationship between the rates of consumption:
* Let the rate of disappearance (consumption) of $\text{X}$ be defined as:
$$\text{Rate of consumption of X} = -\frac{d[\text{X}]}{dt}$$
* Let the rate of disappearance (consumption) of $\text{Y}$ be defined as:
$$\text{Rate of consumption of Y} = -\frac{d[\text{Y}]}{dt}$$
* Based on the stoichiometric coefficients, the rate of consumption of $\text{Y}$ is always twice the rate of consumption of $\text{X}$:
$$-\frac{d[\text{Y}]}{dt} = 2 \times \left(-\frac{d[\text{X}]}{dt}\right)$$
4. Apply this stoichiometry to Trial 2:
Stoichiometric relationships between rates of consumption are determined strictly by the balanced chemical equation and apply to any point in any trial of the reaction, regardless of the initial starting concentrations of the reactants. Therefore, in Trial 2, $\text{Y}$ will be consumed twice as rapidly as $\text{X}$.
5. Conclude the correct option: This identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: While it is true that in Trial 2, $\text{X}$ has a higher starting concentration ($1.00\text{ M}$) than $\text{Y}$ ($0.50\text{ M}$), initial concentration does not dictate the relative rate of consumption. Stoichiometry dictates that $\text{Y}$ is consumed twice as fast as $\text{X}$ regardless of which reactant is present in a higher initial concentration.
• Option B is incorrect: This option incorrectly claims that $\text{X}$ is consumed more rapidly and cites the reaction order. Let's find the reaction order with respect to $\text{X}$:
• From Trial 1 to Trial 2, $[\text{Y}]$ is held constant at $0.50\text{ M}$ while $[\text{X}]$ doubles from $0.50\text{ M}$ to $1.00\text{ M}$. The rate of formation of $\text{XY}_2$ quadruples: $\frac{3.2 \times 10^{-2}}{8.0 \times 10^{-3}} = 4$. Because $2^2 = 4$, the reaction is indeed second order with respect to $\text{X}$.
• However, the reaction order (the exponent in the rate law) describes how changing the concentration of a reactant affects the overall rate of the reaction. It has no bearing on the relative rate of consumption of $\text{X}$ versus $\text{Y}$ in a single ongoing reaction, which is governed solely by stoichiometry.
• Option C is incorrect: This option incorrectly claims that $\text{Y}$ is consumed more rapidly *because* the reaction is second order with respect to $\text{Y}$. Let's check the reaction order with respect to $\text{Y}$:
• From Trial 2 to Trial 3, $[\text{X}]$ is held constant at $1.00\text{ M}$ while $[\text{Y}]$ doubles from $0.50\text{ M}$ to $1.00\text{ M}$. The rate of formation of $\text{XY}_2$ doubles: $\frac{6.4 \times 10^{-2}}{3.2 \times 10^{-2}} = 2$. Because $2^1 = 2$, the reaction is actually first order with respect to $\text{Y}$.
• More importantly, as with Option B, reaction order does not govern relative rates of consumption in an active mixture; stoichiometric coefficients do.