A) + 37.56
B) - 43.76
C) + 43.76
D) + 40.66
Correct Answer: A
Solution :
\[{{H}_{2}}O(l)\xrightarrow[{}]{{{100}^{o}}C}{{H}_{2}}O(g)\] \[{{\Delta }_{vap}}{{H}^{\Theta }}={{\Delta }_{vap}}{{E}^{\Theta }}+\Delta {{n}_{g}}RT\] For the above reaction, \[\Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}=1-0=1\] \[\therefore \]\[40.66\,kJ\,mo{{l}^{-1}}\,={{\Delta }_{vap}}{{E}^{\Theta }}+1\times 8.314\] \[\times {{10}^{-3}}\times 373\] \[{{\Delta }_{vap}}{{E}^{\Theta }}=40.66kJ\,mo{{l}^{-1}}-3.1\,kJ\,mo{{l}^{-1}}\] \[=+37.56\,kJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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