Answer:
The equation to show the the correct form to show the standard molar enthalpy of formation:
[tex]\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ[/tex]
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Given, that 1 mole of [tex]H_2[/tex] gas and 1 mole of [tex]Br_2[/tex] liquid gives 2 moles of HBr gas as a product.The reaction releases 72.58 kJ of heat.
[tex]H_2(g) + Br_2(l)\rightarrow 2HBr(g) ,\Delta H_{f}^o= -72.58kJ[/tex]
Divide the equation by 2.
[tex]\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ[/tex]
The equation to show the the correct form to show the standard molar enthalpy of formation:
[tex]\frac{1}{2}H_2(g) +\frac{1}{2}Br_2(l)\rightarrow HBr(g) ,\Delta H_{f}^o= -36.29 kJ[/tex]