| A[M] | B[M] | initial rate of formation of D | |
| I | 0.1 | 0.1 | 6.0 × 10-3 |
| II | 0.3 | 0.2 | 7.2 × 10-2 |
| III | 0.3 | 0.4 | 2.88 × 10-1 |
| IV | 0.4 | 0.1 | 2.40 × 10-2 |
From the rate law:
\[ r = k[A]^x[B]^y \]
Using the data:
\[ \text{From (I) and (IV): } \frac{2.4 \times 10^{-2}}{6 \times 10^{-3}} = \frac{(0.4)^x}{(0.1)^x} \implies 4 = 4^x \implies x = 1 \]
\[ \text{From (III) and (II): } \frac{2.88 \times 10^{-1}}{7.2 \times 10^{-2}} = \frac{(0.4)^y}{(0.2)^y} \implies 4 = 2^y \implies y = 2 \]
Overall order = \(x + y = 1 + 2 = 3\).
Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below:
Based on the above data, predict the order of the reaction and write the expression for the rate law.
Let \( C_{t-1} = 28, C_t = 56 \) and \( C_{t+1} = 70 \). Let \( A(4 \cos t, 4 \sin t), B(2 \sin t, -2 \cos t) \text{ and } C(3r - n_1, r^2 - n - 1) \) be the vertices of a triangle ABC, where \( t \) is a parameter. If \( (3x - 1)^2 + (3y)^2 = \alpha \) is the locus of the centroid of triangle ABC, then \( \alpha \) equals:
A line charge of length \( \frac{a}{2} \) is kept at the center of an edge BC of a cube ABCDEFGH having edge length \( a \). If the density of the line is \( \lambda C \) per unit length, then the total electric flux through all the faces of the cube will be : (Take \( \varepsilon_0 \) as the free space permittivity)