MCQ
Each pair forms ideal solution except
- A$C_2H_5Br$ and $C_2H_5I$
- B$C_6H_5Cl$ and $C_6H_5Br$
- C$C_6H_6$ and $C_6H_5CH_3$
- ✓$C_2H_5I$ and $C_2H_5OH$
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In which case the value of $\Delta H_{sol.} < 0$
| Rate Constant | Activation energy | |
| Step $1$ | $k_1$ | $E_{a_1} = 180\ kJ/mol$ |
| Step $2$ | $k_2$ | $E_{a_2} = 80\ kJ/mol$ |
| Step $3$ | $k_3$ |
$E_{a_3} = 50\ kJ/mol$ |


Reason : Low spin complexes have lesser number of unpaired electrons.