Rate of reaction is given by the change in concentration of FeSO₄: \[ \frac{-\Delta [FeSO₄]}{\Delta t} \] Substitute the given values: \[ \frac{-10 + 8.8}{30 \times 60} = \frac{1.2}{1800} = 6.67 \times 10^{-4} \] From the given reaction, the rate of production of Fe₂(SO₄)₃ is related to the rate of FeSO₄: \[ \frac{1}{6} \times \frac{-\Delta [FeSO₄]}{\Delta t} \] Substitute the value of \(\frac{-\Delta [FeSO₄]}{\Delta t}\): \[ \text{Rate of production of Fe}_2(SO₄)_3 = \frac{3}{6} \times 6.67 \times 10^{-4} = 333.33 \times 10^{-6} \] Thus, the rate of production of Fe₂(SO₄)₃ is \(333 \times 10^{-6}\) mol L⁻¹ s⁻¹. Hence,
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.
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.
In the circuit shown, assuming the threshold voltage of the diode is negligibly small, then the voltage \( V_{AB} \) is correctly represented by:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: