A) -13.0 kcal
B) -27.0 kcal
C) 27.0 kcal
D) 13.0 kcal
Correct Answer: C
Solution :
Key Idea The relation between \[\Delta H\] and is \[\Delta E\] given by \[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] where \[\Delta {{n}_{g}}=\] change in the number of gaseous moles. Vaporisation of 3 moles of water is \[3{{H}_{2}}O(l)\xrightarrow{{}}3{{H}_{2}}O(g)\] \[\Delta {{n}_{g}}=3-0=3\] Latent heat of vaporisation =10 kcal/mol So, heat change for 3 moles of water to vapours \[=3\times 10=30\] kcal Here, T = 500 K R = 0.002 kcal \[mo{{l}^{-1}}{{K}^{-1}}\] \[\Delta H=30\] kcal Put these value in the above relation \[30=\Delta E+3\times 0.002\times 500\] \[30=\Delta E+3\] \[\Delta E=30-3=27\] kcal
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