Question

Decomposition of $HI(g)$ on Gold surface is zero order reaction. Initially, few moles of $H_2$ are present in container then which of the following graph is correct?

• A. Graph A: A plot of $P_{H_2}$ versus time. The y-axis is labeled $P_{H_2}$ and the x-axis is labeled time. The graph shows a horizontal line, indicating that the partial pressure of $H_2$ remains constant over time.
• B. Graph B: A plot of $P_{H_2}$ versus time. The y-axis is labeled $P_{H_2}$ and the x-axis is labeled time. The graph shows a straight line with a positive slope, indicating that the partial pressure of $H_2$ increases linearly with time, starting from a non-zero initial pressure.
• C. Graph C: A plot of $P_{H_2}$ versus time. The y-axis is labeled $P_{H_2}$ and the x-axis is labeled time. The graph shows a straight line with a positive slope, indicating that the partial pressure of $H_2$ increases linearly with time, starting from zero initial pressure.
• D. Graph D: A plot of $P_{H_2}$ versus time. The y-axis is labeled $P_{H_2}$ and the x-axis is labeled time. The graph shows a straight line with a negative slope, indicating that the partial pressure of $H_2$ decreases linearly with time.

Answer 1

Answer: (B)

Graph B: A plot of $P_{H_2}$ versus time. The y-axis is labeled $P_{H_2}$ and the x-axis is labeled time. The graph shows a straight line with a positive slope, indicating that the partial pressure of $H_2$ increases linearly with time, starting from a non-zero initial pressure.

Answer 2

For a zero order reaction, the rate (rate = k) is constant and independent of reactant concentration. Thus, the amount of product (here, H₂) formed increases linearly with time according to:

$P_{H_2} = P_{H_2}^0 + kt$

Since there is an initial amount of H₂ present (i.e., $P_{H_2}^0 \neq 0$), the graph will start from a non-zero value and increase linearly. This matches Graph B.