For a cyclic process, the change in internal energy (\(\Delta U\)) is zero. Therefore, the heat absorbed (\(\Delta Q\)) is equal to the work done (\(\Delta W\)), which is the area enclosed by the P-V curve. The area is calculated as:
\[ \Delta Q = \Delta W = \pi \times (140 \times 10^3) \, \text{Pa} \times (140 \times 10^{-6} \, \text{m}^3) \]
\[ \Delta Q = 61.6 \, \text{J} \]
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
\([A]\) (mol/L) | \(t_{1/2}\) (min) |
---|---|
0.100 | 200 |
0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.