The enthalpy of the reaction forming $\mathrm{PbO}$ according to the following equation is $438…

The enthalpy of the reaction forming $\mathrm{PbO}$ according to the following equation is $438 \mathrm{~kJ}$. What heat energy $(\mathrm{kJ})$ is released in formation of $22.3 \mathrm{~g} \mathrm{PbO}(\mathrm{s}) ?$
(Atomic masses : $\mathrm{Pb}=207, \mathrm{O}=16.0$ )
$2 \mathrm{~Pb}(\mathrm{~s})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{PbO}(\mathrm{s})$
  1. $21.9$
  2. $28.7$
  3. $14.6$
  4. $34.2$

Solution

$\mathrm{Q}=\frac{1}{2} 438 \times \frac{22.3}{223}=21.9 \mathrm{~kJ}$ .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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