Chemical Thermodynamics Question 233
Question: The enthalpy of neutralisation of $ NH_4OH $ and $ CH_3COOH $ is $ -10.5kcalmo{{l}^{-1}} $ and enthalpy of neutralisation of $ CH_3COOH $ with strong base is $ -12.5,kcalmo{{l}^{-1}} $ . The enthalpy of ionisation of $ NH_4OH $ will be
Options:
A) $ 3.2\ \text{kcal}\ \text{mol}^{-1}$
B) $ 2.0\ \text{kcal}\ \text{mol}^{-1}$
C) $ 3.0\ \text{kcal}\ \text{mol}^{-1}$
D) $ 4.0\ \text{kcal}\ \text{mol}^{-1}$
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Answer:
Correct Answer: B
Solution:
$ {\Delta_{neu}}H $ for strong base and strong acid $ =-13.7kcal,mol^{-1}$
$ \Delta H_{ion}(CH_3COOH) $
$ =-12.5-( -13.7 )=1.2\text{kcal}\cdot\text{mol}^{-1}$
$ \Delta H_{ion}(NH_4OH) $
$ =-10.5-(-13.7)-\Delta H_{ion}(CH_3COOH) $
$ =13.7-10.5-1.2 $
$ =2,kcal,mol^{-1}$
BETA
