Question

How much heat should be transferred when 20.1 g of hydrogen bromide is formed during the reaction of hydrogen gas with liquid bromine?

Answer 1

Solution.

$m(HBr)=20.1g;$

$M(HBr)=81g/mol;$

$\nu(HBr)=\dfrac{m(HBr)}{M(HBr)};$

$\nu(HBr)=\dfrac{20.1g}{81g/mol}=0.248mol;$

The enthalpy of formation of hydrogen bromide is equal to -34.1kJ/mol, that is, in the formation of one mole of hydrogen bromide, 34 kJ of heat is absorbed.

$1mol----34kJ;$

$0.248mol--xkJ;$

$x=\dfrac{0.248mol\sdot34kJ}{1mol}=8.437kJ;$

Answer: 8.437kJ of heat is absorbed.