Chemical Thermodynamics Question 261
Question: $ \Delta H_{f}^{0} $ of water is $ -285.8kJmo{{l}^{-1}} $ . If enthalpy of neutralisation of monoacidic strong base is $ -57.3kJmo{{l}^{-1}} $ . $ \Delta H_{f}^{0} $ of $ O{{H}^{-}} $ ion will be
Options:
A) $ -114.25kJ,mo{{l}^{-1}} $
B) $ 114.25kJmo{{l}^{-1}} $
C) $ 228.5kJ,mo{{l}^{-1}} $
D) $ -228.5kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ H_2(g)+\frac{1}{2}O_2(g)\xrightarrow{{}}H_2O(l); $
$ \Delta H=-285.8kJ $ …. (i) $ {{H}^{+}}(aq)+O{{H}^{-}}(aq)\xrightarrow{{}}H_2O(l); $
$ \Delta H=-57.3kJ $ …. (ii) $ \frac{1}{2}H_2(g)+aq\xrightarrow{{}}{{H}^{+}}{{(aq)}^{+}}{{e}^{-}};\Delta H=0 $ (by convention) …. (iii) (ii) - (iii) gives, $ \frac{1}{2}H_2( g )+\frac{1}{2}O_2( g )+{{e}^{-}}+aq\xrightarrow{{}}O{{H}^{-}}( aq ) $
$ \Delta H=-285.8+57.3=-228.5kJ $
BETA
