The torque \(\tau\) is given by:
\[ \tau = pE \sin \theta \] This can be written as: \[ \tau = (2aq) E \sin \theta \] Substituting the known values: \[ \tau = \left(5 \times 10^{-3} \times 1 \times 10^{-12} \times 10^3\right) \times \frac{4}{5} \] Simplifying: \[ \tau = 4 \times 10^{-12} \, \text{Nm} \] The direction of the torque is along the negative \(Z\)-direction.
A certain reaction is 50 complete in 20 minutes at 300 K and the same reaction is 50 complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. Given: \[ R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1}, \quad \log 4 = 0.602 \]