Question

Calculate the standard enthalpy of the 3rd reaction using the given data: H2(g) +126) +2 HIO) A,Hº = +52.96 kJ/mol 2 H2(g) + O2(g) + 2 H209) A,Hº = - 483.64 kJ/mol 4 HI) + O2(g) +2 12(8) + 2 H206) A,Hº =? A.H=-589.5619/mol Calculate the internal energy change for reaction (3) in the previous problem.

Answer 1

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.

Step-by-step explanation:

`H_2(g) +I_2(s) \rightarrow 2 HI(g) ,\Delta H^(o)_(r,1)= +52.96 kJ/mol`...[1]

`2 H_2(g) + O_2(g)\rightarrow 2 H_2O(g),\Delta H^(o)_(r,2)=-483.64 kJ/mol`...[2]

`4 HI(g)+O_2(g)\rightarrow 2 I_2(s)+2 H_2O(g) ,\Delta H^(o)_(r,3) =?`..[3]

By using Hess's we can determine the standard enthalpy of reaction of third reaction:

[2] - 2 × [1] = [3] (Hess's law)

`O_2+4HI\rightarrow 2H_2O(g)+2I_2(s)`

`\Delta H^(o)_(r,3)=\Delta H^(o)_(r,2)-2* \Delta H^(o)_(r,1)`

`=-483.64 kJ/mol - 2* (52.96 kJ/mol)=-586.56 kJ/mol`

The negative sign indicates that energy is released during this reaction.

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.