Catalysts lower activation energies equally for forward and backward reactions without altering ∆H
Step 1: Activation energy for the uncatalyzed backward reaction:
\( E_a(\text{backward}) = E_a(\text{forward}) - \Delta H \)
\( E_a(\text{backward}) = 300 - 20 = 280 \, \text{kJ/mol} \)
Step 2: Using the given temperatures and equal rates, calculate \( E_a(\text{forward, catalyzed}) \):
\( \frac{E_a(\text{forward, catalyzed})}{E_a(\text{forward, uncatalyzed})} = \frac{T_c}{T_u} \)
\( \frac{E_a(\text{forward, catalyzed})}{300} = \frac{300}{600} \)
\( E_a(\text{forward, catalyzed}) = 150 \, \text{kJ/mol} \)
Step 3: Calculate \( E_a(\text{backward, catalyzed}) \):
\( E_a(\text{backward, catalyzed}) = E_a(\text{forward, catalyzed}) - \Delta H \)
\( E_a(\text{backward, catalyzed}) = 150 - 20 = 130 \, \text{kJ/mol} \)
The colour of the solution observed after about 1 hour of placing iron nails in copper sulphate solution is:
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 