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:
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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: Read the two reactions. Experiment I is $\text{X(g)} \rightleftharpoons 2\text{Y(g)}$ and Experiment II is $\text{X(g)} \rightleftharpoons \text{Z(g)}$. Both run at fixed temperature and volume, so partial pressures behave like amounts. We want how the reactant pressure changes as the product pressure changes.
Step 2: Track Experiment II first. In $\text{X} \rightleftharpoons \text{Z}$, every unit of X that disappears makes exactly one unit of Z. So if $p_Z$ rises by some amount, $p_X$ falls by the same amount. That gives a straight line with slope $-1$.
Step 3: Write Experiment II as an equation. If $p_0$ is the starting pressure of X, then at any time $p_X + p_Z = p_0$, so $p_X = p_0 - p_Z$. This is a straight line of slope minus one when we plot $p_X$ against $p_Z$.
Step 4: Track Experiment I. In $\text{X} \rightleftharpoons 2\text{Y}$, one unit of X that reacts produces two units of Y. So for every unit of Y formed, only half a unit of X is lost. That means $p_X$ falls only half as fast as $p_Y$ rises.
Step 5: Write Experiment I as an equation. Here $p_X = p_0 - \tfrac{1}{2}p_Y$, which is again a straight line but with a gentler slope of $-\tfrac{1}{2}$. So Experiment I is a line less steep than Experiment II.
Step 6: Pick the matching plot. So both plots are straight lines starting at the same reactant pressure, with Experiment II steeper (slope $-1$) and Experiment I gentler (slope $-\tfrac{1}{2}$). The option showing these two straight lines with those slopes is correct, which is option A.
\[ \boxed{\text{Two straight lines: slope } -1 \text{ for II and } -\tfrac{1}{2} \text{ for I (option A)}} \]
