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Question
according to the graph, what happens to the concentration of a over time compared to a₂? reaction: 2a ⇌ a₂ graph: concentration (m) vs time (sec), a starts high and decreases, a₂ starts at 0 and increases options: a increases at the same rate as a₂ decreases. a decreases faster than a₂ increases. a decreases at the same rate as a₂ increases. a increases faster than a₂ decreases.
- First, analyze the reaction: \( 2A
ightleftharpoons A_2 \). This means that 2 moles of A react to form 1 mole of \( A_2 \).
- Now, look at the graph: The concentration of A (purple line) starts high and decreases over time. The concentration of \( A_2 \) (red line) starts at 0 and increases over time.
- Compare the rates: From the graph, the slope of the A line (decreasing) is steeper initially than the slope of the \( A_2 \) line (increasing). But we also have to consider the stoichiometry. However, visually, the decrease in A's concentration is faster than the increase in \( A_2 \)'s concentration? Wait, no, wait. Wait, the reaction is \( 2A
ightarrow A_2 \) (forward). So for every 2 moles of A that disappear, 1 mole of \( A_2 \) appears. But looking at the graph, the initial drop of A is steeper than the rise of \( A_2 \), but let's check the options. Wait, the options: Let's re - evaluate.
Wait, the options:
- Option 1: A increases at the same rate as \( A_2 \) decreases. But A is decreasing, \( A_2 \) is increasing. So this is wrong.
- Option 2: A decreases faster than \( A_2 \) increases. Let's see the graph. The purple line (A) has a steeper slope (more rapid change) than the red line (\( A_2 \)) initially. So the rate of decrease of A is faster than the rate of increase of \( A_2 \).
- Option 3: A decreases at the same rate as \( A_2 \) increases. But due to stoichiometry (2 moles of A for 1 mole of \( A_2 \)) and the graph's slope, this is not the case.
- Option 4: A increases faster than \( A_2 \) decreases. A is decreasing, \( A_2 \) is increasing, so this is wrong.
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B. A decreases faster than \( A_2 \) increases.