Order with respect to A
For a first-order reaction:
\[t_{75\%} = 2 \times t_{50\%}.\]
This is consistent with the information given, so the reaction is first order with respect to A.
Order with respect to B The plot of [B] versus $t$ is a straight line, which indicates that the reaction is zero order with respect to B.
Overall order of the reaction:
\[\text{Order} = 1 \, (\text{w.r.t. A}) + 0 \, (\text{w.r.t. B}) = 1.\]
Final Answer:\[1.\]
The following data were obtained for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2N_2O(g) \] at different concentrations:
The rate law of this reaction is:
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} $