Question

9. Calculate the standard enthalpy of the 3rd reaction using the given data: A,H° = +52.96 kJ/mol AFH= -483.64 kJ/mol H2s)+I202 HI 2 H2g)+O2ig)2 H2O() 4 HIg+O2lg)2 12)+2 H2O(e (1) (2) (3) A,H° =? 10. Calculate the internal energy change for reaction (3) in the previous problem.

Answer 1

Answer:

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

Explanation:

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

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

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

The unknown standard enthalpy of third reaction can be calculated by using Hess's law:

The law states that 'the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps'.

[2] - 2 × [1] = [3]

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

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

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

The standard enthalpy of the 3rd reaction is -586.56 kJ/mol.The negative sign indicates that energy is released during this reaction.