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,
In the given graph, \( E_a \) for the reverse reaction will be
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: