We know that 350.5kJ of energy is released by the system. Hence, $deltaH^{\circ}_f$deltaH?ƒ =-350.5kJ/mol, since that for every mole of zinc oxide formed, 350.5kJ leaves the system.
The molar mass of zinc oxide is: 65.38g/mol + 16.00 g/mol = 81.38 g/mol
6 years ago
Answered By Eric C
From the equation:
$Zn\left(s\right)+\frac{1}{2}O_2\left(g\right)->ZnO\left(s\right)+350.5kJ$Zn(s)+12 O2(g)−>ZnO(s)+350.5kJ
We know that 350.5kJ of energy is released by the system. Hence, $deltaH^{\circ}_f$deltaH?ƒ =-350.5kJ/mol, since that for every mole of zinc oxide formed, 350.5kJ leaves the system.
The molar mass of zinc oxide is: 65.38g/mol + 16.00 g/mol = 81.38 g/mol
Therefore,
$n_{ZnO\left(s\right)}=\frac{25.0g}{\frac{81.38g}{mol}}=0.307mol$nZnO(s)=25.0g81.38gmol =0.307mol
From this we can solve for the standard enthalpy of formation:
$H^o=-\frac{350.5kJ}{mol}\times0.3072mol=-107.7kJ$Ho=−350.5kJmol ×0.3072mol=−107.7kJ