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

Answer:

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

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)

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

$\Delta H^{o}_{r,3}=\Delta H^{o}_{r,2}-2\times \Delta H^{o}_{r,1}$

$=-483.64 kJ/mol - 2\times (52.96 kJ/mol)=-586.56 kJ/mol$

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

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