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 = \(x + y = 1 + 2 = 3\).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32