The reaction is: \[ \text{A(g)} \rightarrow 2\text{B(g)} + \text{C(g)} \]
Given:
\(P_{23} = P_0 + 2x = 200 \\ P_\infty = 3P_0 = 300 \\ P_0 = 100\)
The rate constant $K$ is calculated using:
\[ K = \frac{1}{t} \ln \frac{P_\infty - P_0}{P_\infty - P_t} \]
Substituting the values:
\[ K = \frac{2.3}{23} \log \frac{300 - 100}{300 - 200} \] \[ K = \frac{2.3 \times 0.301}{23} = 0.0301 = 3.01 \times 10^{-2} \, \text{s}^{-1} \]
The correct answer is (3).
In the given graph, \( E_a \) for the reverse reaction will be
The velocity-time graph of an object moving along a straight line is shown in the figure. What is the distance covered by the object between \( t = 0 \) to \( t = 4s \)?
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $