The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol$-$1, while the electron gain enthalpy of Br is $-$325.0 kJ mol$-$1. Given the lattice enthalpy of NaBr is $-$728.4 kJ mol$-$1. The energy for the formation of NaBr ionic solid is ($-$) ____________ $\times$ 10$-$1 kJ mol$-$1.
Answer (integer)
5576
Solution
$Na(s)\buildrel {} \over
\longrightarrow N{a^ + }(g)$<br><br>$\Delta H = 495.8$<br><br>${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over
\longrightarrow B{r^ - }(g)$<br><br>$\Delta H = 325$<br><br>$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over
\longrightarrow NaBr(s)$<br><br>$\Delta H = - 728.4$<br><br>$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over
\longrightarrow NaBr(s).$<br><br>$\Delta H = ?$<br><br>$\Delta H = 495.8 - 325 - 728.4 - 557.6$ kJ<br><br>$= - 5576 \times {10^{ - 1}}$ kJ
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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