JEE MAIN - Chemistry (2021 - 25th February Morning Shift - No. 17)
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
5576
Explanation
$$Na(s)\buildrel {} \over
\longrightarrow N{a^ + }(g)$$
$$\Delta H = 495.8$$
$${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over \longrightarrow B{r^ - }(g)$$
$$\Delta H = 325$$
$$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over \longrightarrow NaBr(s)$$
$$\Delta H = - 728.4$$
$$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over \longrightarrow NaBr(s).$$
$$\Delta H = ?$$
$$\Delta H = 495.8 - 325 - 728.4 - 557.6$$ kJ
$$ = - 5576 \times {10^{ - 1}}$$ kJ
$$\Delta H = 495.8$$
$${1 \over 2}B{r_2}(l) + {e^ - }\buildrel {} \over \longrightarrow B{r^ - }(g)$$
$$\Delta H = 325$$
$$N{a^ + }(g) + B{r^ - }(g)\buildrel {} \over \longrightarrow NaBr(s)$$
$$\Delta H = - 728.4$$
$$Na(s) + {1 \over 2}B{r_2}(l)\buildrel {} \over \longrightarrow NaBr(s).$$
$$\Delta H = ?$$
$$\Delta H = 495.8 - 325 - 728.4 - 557.6$$ kJ
$$ = - 5576 \times {10^{ - 1}}$$ kJ
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