A) \[160.3\]
B) \[173.4\]
C) \[868.3\]
D) \[672.5\]
Correct Answer: B
Solution :
\[L{{i}_{(g)}}\,\xrightarrow{\,}\,Li_{(g)}^{+}\,+{{e}^{-}}\] \[\Delta {{H}_{1}}=5.41\,\times 1.6\times {{10}^{-22}}kJ/mole\] \[=5.41\,\times 1.6\,\times {{10}^{-22}}\times 6.02\,\times {{10}^{23}}kJ/mole\] \[C{{\ell }_{(g)}}\,+{{e}^{-}}\xrightarrow{\,}\,C{{\ell }^{-}}_{(g)}\] \[\Delta {{H}_{2}}=-3.61\,\times 1.6\times {{10}^{-22}}\,kJ/mole\] \[=-3.61\,\times 1.6\,\times {{10}^{-22}}\,\times 6.02\,\times {{10}^{23}}\,kJ/mole\] on adding \[L{{i}_{(g)}}+C{{l}_{(g)}}\xrightarrow{\,}\,Li_{(g)}^{+}+C{{\ell }^{-}}_{(g)}\] \[=\Delta {{H}_{2}}=\Delta {{H}_{1}}+\Delta {{H}_{2}}=173.4\,\,kJ/mole\]You need to login to perform this action.
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