Consider the following chemical reactions performed at identical temperature and volume:
Experiment I: $\text{X(g)} \rightleftharpoons 2\text{Y(g)}$
Experiment II: $\text{X(g)} \rightleftharpoons \text{Z(g)}$
The partial pressure of the reactant and product are denoted, respectively, by $p_{\text{reactant}}$ and $p_{\text{product}}$ during the course of the reactions. Assuming ideal gas behaviour, the correct plot of $p_{\text{reactant}}$ versus $p_{\text{product}}$ for Experiments I and II is:
Show Hint
Stoichiometry dictates the rate of change of pressures. Since one mole of X produces two moles of Y, the pressure of the product in Experiment I grows twice as fast as it does in Experiment II, making the slope of $p_{\text{reactant}}$ vs $p_{\text{product}}$ half as steep.
Step 1: Understanding the Question:
The question asks us to identify the correct graphical representation of the partial pressure of the reactant ($p_{\text{reactant}}$) versus the partial pressure of the product ($p_{\text{product}}$) for two different gas-phase dissociation experiments.
Step 2: Key Formula or Approach:
For gas-phase reactions at constant volume and temperature, the change in partial pressure of any component is directly proportional to its stoichiometric coefficient:
\[ \Delta P \propto \text{stoichiometric coefficient} \]
Step 3: Detailed Explanation:
• Let the initial partial pressure of the reactant gas X in both experiments be $P_0$.
• Let us analyze Experiment I: $\text{X(g)} \rightleftharpoons 2\text{Y(g)}$.
• If at any point during the reaction, a pressure $x$ of reactant X has decomposed:
\[ p_{\text{reactant}} = P_0 - x \]
\[ p_{\text{product}} = 2x \implies x = \frac{p_{\text{product}}}{2} \]
• Substituting this value of $x$ into the reactant equation:
\[ p_{\text{reactant}} = P_0 - 0.5\ p_{\text{product}} \]
• This is a straight line equation with a y-intercept of $P_0$ and a slope of $-0.5$.
• Let us analyze Experiment II: $\text{X(g)} \rightleftharpoons \text{Z(g)}$.
• If at any point during the reaction, a pressure $y$ of reactant X has decomposed:
\[ p_{\text{reactant}} = P_0 - y \]
\[ p_{\text{product}} = y \]
• Substituting this value of $y$ into the reactant equation:
\[ p_{\text{reactant}} = P_0 - p_{\text{product}} \]
• This is a straight line equation with a y-intercept of $P_0$ and a slope of $-1.0$.
• Comparing the two plots:
itemize
• Both plots are straight lines that originate from the exact same initial value ($P_0$) on the vertical axis.
• The slope for Experiment II ($-1.0$) is twice as steep as the slope for Experiment I ($-0.5$).
itemize
Step 4: Final Answer:
Therefore, the correct graphical representation is the one shown in Option A.
