Question

Hydrogen peroxide can be prepared in several ways. One method is the reaction between hydrogen and oxygen, another method is the reaction between water and oxygen. Calculate the ?G�rxn of each reaction below using values from this table.
1.) H2(g) + O2(g) ---> H2O2(l)
<---
Delta G rxn = ? KJ.mol-1
2.) H2O(l) +1/2O2 (g) ---> H2O2(l)
<----
Delta G rxn = ? KJ.mol-1
3. Which method requires less energy under standard conditions?
Reaction 1 or Reaction 2

Answer 1

Answer :

(1) The values of $(\Delta G_{rxn})$ is, -120.4 kJ/mol

(2) The values of $(\Delta G_{rxn})$ is, 116.7 kJ/mol

Explanation :

(1) The given balanced chemical reaction is,

$H_2(g)+O_2(g)\rightarrow H_2O_2(l)$

Now we have to calculate the $(\Delta G_{rxn})$.

$\Delta G_{rxn}=G_f_{product}-G_f_{reactant}$

$\Delta G_{rxn}=[n_{H_2O_2(l)}\times \Delta G_f^0_{(H_2O_2(l))}]-[n_{H_2(g)}\times \Delta G_f^0_{(H_2(g))}+n_{O_2(g)}\times \Delta G_f^0_{(O_2(g))}]$

where,

$\Delta G_{rxn}$ = enthalpy of reaction = ?

n = number of moles

$\Delta G_f^0_{(H_2O_2(l))}=-120.4kJ/mol$

$\Delta G_f^0_{(H_2(g))}=0kJ/mol$

$\Delta G_f^0_{(O_2(g))}=0kJ/mol$

Now put all the given values in this expression, we get:

$\Delta G_{rxn}=[1mole\times (-120.4kJ/mol)]-[1mole\times (0kJ/mol)+1mole\times (0kJ/mol)]$

$\Delta G_{rxn}=-120.4kJ/mol$

(2) The given balanced chemical reaction is,

$H_2O(l)+\frac{1}{2}O_2(g)\rightarrow H_2O_2(l)$

Now we have to calculate the $(\Delta G_{rxn})$.

$\Delta G_{rxn}=G_f_{product}-G_f_{reactant}$

$\Delta G_{rxn}=[n_{H_2O_2(l)}\times \Delta G_f^0_{(H_2O_2(l))}]-[n_{H_2O(l)}\times \Delta G_f^0_{(H_2O(l))}+n_{O_2(g)}\times \Delta G_f^0_{(O_2(g))}]$

where,

$\Delta G_{rxn}$ = enthalpy of reaction = ?

n = number of moles

$\Delta G_f^0_{(H_2O_2(l))}=-120.4kJ/mol$

$\Delta G_f^0_{(H_2O(l))}=-237.1kJ/mol$

$\Delta G_f^0_{(O_2(g))}=0kJ/mol$

Now put all the given values in this expression, we get:

$\Delta G_{rxn}=[1mole\times (-120.4kJ/mol)]-[1mole\times (-237.1kJ/mol)+\frac{1}{2}mole\times (0kJ/mol)]$

$\Delta G_{rxn}=116.7kJ/mol$