Hi,
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. Given the ?G°rxn values for each reaction, which method (reaction 1 or 2) requires less energy under standard conditions?
1. H2(gas) + O2(gas) <---> H2O2(liquid) ?G°rxn = -120.4 kJ
2. H2O(liquid) + 1/2 O2(gas) <---> H2O2(liquid) ?G°rxn = +116.7 kJ
Would the answer be reaction 2 because it is lower in magnitude? I know that a negative free energy value indicates a spontaneous reaction and a positive value a nonspontaneous reaction but is the sign significant in this case?
?G = ?G° + RTln(Q)
?G = -72.6 kJ + (0.008314 kJ mol- K-)(298 K)(ln 32.4)
?G = -72.6 + 8.6174 = -63.98 kJ
I am not sure about my approach to this question. Can anyone confirm if I am on the right track?
Thanks. All help Is very much appreciated.
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. Given the ?G°rxn values for each reaction, which method (reaction 1 or 2) requires less energy under standard conditions?
1. H2(gas) + O2(gas) <---> H2O2(liquid) ?G°rxn = -120.4 kJ
2. H2O(liquid) + 1/2 O2(gas) <---> H2O2(liquid) ?G°rxn = +116.7 kJ
Would the answer be reaction 2 because it is lower in magnitude? I know that a negative free energy value indicates a spontaneous reaction and a positive value a nonspontaneous reaction but is the sign significant in this case?
?G = ?G° + RTln(Q)
?G = -72.6 kJ + (0.008314 kJ mol- K-)(298 K)(ln 32.4)
?G = -72.6 + 8.6174 = -63.98 kJ
I am not sure about my approach to this question. Can anyone confirm if I am on the right track?
Thanks. All help Is very much appreciated.