For first order reaction: 2A(g) → 4B(g) + C(g) Total pressure at \( t = 30 \, \text{sec} \) and \( t = \infty \) are 300 torr and 600 torr respectively. Calculate pressure of C(g) at 30 sec (in torr).
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For a first-order reaction, the total pressure is related to the change in the number of moles of gas. Use stoichiometry to relate the change in pressure to the amount of substance reacted.
Step 1: Understanding the reaction.
The given first-order reaction is:
\[
2A(g) \rightarrow 4B(g) + C(g)
\]
This indicates that when 2 moles of A react, 4 moles of B and 1 mole of C are produced. Therefore, the total number of moles of products formed is proportional to the progress of the reaction.
Step 2: Relating pressure to the number of moles.
At constant temperature and volume, the pressure of a gas is directly proportional to the number of moles present. Hence, the total pressure of the system changes as the reaction proceeds.
Given:
- Total pressure at \( t = 30 \, \text{sec} \) is \( 300 \, \text{torr} \)
- Total pressure at \( t = \infty \) is \( 600 \, \text{torr} \) (reaction completion)
Thus, the fraction of reaction completed at 30 seconds is:
\[
\text{Fraction completed} = \frac{P_{30}}{P_{\infty}} = \frac{300}{600} = \frac{1}{2}
\]
Step 3: Determining the pressure contribution of gas C.
From the reaction stoichiometry:
\[
2A \rightarrow 4B + C
\]
Total moles of gaseous products formed = \( 4 + 1 = 5 \)
Out of these 5 parts, the contribution of gas C corresponds to 1 part.
Therefore, the pressure of gas C at \( t = 30 \, \text{sec} \) is:
\[
P_C = \frac{1}{5} \times 300
\]
\[
P_C = 60 \, \text{torr}
\]
Final Answer: \( P_C = 60 \, \text{torr} \).
