Question

How much heat is liberated when one mole of gaseous $N{a}^{\oplus }$ combines with one mole of $C{l}^{\ominus }$ ion to form solid $NaCl$.

Use the data given below:
$Na\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)⟶NaCl\left(s\right)$; $\Delta H=-98.23kcal$
$Na\left(s\right)⟶Na\left(g\right)$;$\Delta H=+25.98kcal$
$Na\left(g\right)⟶N{a}^{\oplus }+e^{-}$;$\Delta H=+120.0kcal$
$C{l}_2\left(g\right)⟶2Cl\left(g\right)$;$\Delta H=+58.02kcal$
$C{l}^{\ominus }\left(g\right)⟶Cl\left(g\right)+e^{-}$;$\Delta H=+87.3kcal$

• A. $\Delta H=-185.92Kcal$
• B. $\Delta H=+185.92Kcal$
• C. $\Delta H=0Kcal$
• D. None of these

Answer 1

Answer: (A)

$\Delta H=-185.92Kcal$

$N{a}^{\oplus }\left(g\right)+C{l}^{\ominus }\left(g\right)⟶NaCl\left(s\right)$$;\Delta H=?$

Given:

$Na\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)⟶NaCl\left(s\right)$$;\Delta H=-98.23kcal$

$Na\left(s\right)⟶Na\left(g\right)$ $;\Delta H=25.98kcal$

$Na\left(g\right)⟶N{a}^{\oplus }\left(g\right)+e^{-}$ $;\Delta H=120.0kcal$

$\frac{1}{2}C{l}_2\left(g\right)⟶Cl\left(g\right)$$;\Delta H=58.02\times \frac{1}{2}kcal$

$C{l}^{\ominus }\left(g\right)⟶Cl\left(g\right)+e^{-}$$;\Delta H=87.3kcal$

Rewriting equations:

$Na\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)⟶NaCl\left(s\right)$$;\Delta H_1=-98.23$

$Na\left(g\right)⟶Na\left(s\right)$ $;\Delta H_2=-25.98$

$N{a}^{\oplus }\left(g\right)+e^{-}⟶Na\left(g\right)$ $;\Delta H_3=-120.0$

$\frac{1}{2}\left[2Cl\right(g)⟶C{l}_2(g\left)\right]$$;\Delta H_4=-58.02\times \frac{1}{2}$

$C{l}^{\ominus }\left(g\right)⟶Cl\left(g\right)+e^{-}$$;\Delta H_5=87.3$

$N{a}^{\oplus }\left(g\right)+C{l}^{\ominus }⟶NaCl\left(s\right)$

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$
$=-98.23-25.98-120.0-\frac{1}{2}\times 58.02+87.3$
$=-185.92Kcal$