Asked by melinda
Calculate the standared enthalpy of formation of solid Mg(OH)2 given the following data:
2Mg(s) + O29(g) →2MgO(s) ∆H = -1203.6kJ
Mg(OH)2(s) →MgO(s) + H2O(l) ∆H= +37.1kJ
2H2(g) +O2(g) →2H2O(l) ∆H = -571.7kJ
2Mg(s) + O29(g) →2MgO(s) ∆H = -1203.6kJ
Mg(OH)2(s) →MgO(s) + H2O(l) ∆H= +37.1kJ
2H2(g) +O2(g) →2H2O(l) ∆H = -571.7kJ
Answers
Answered by
DrBob222
Use eqn 1 as is.
Reverse eqn 2 and multiply by 2. (also multiply delta H by 2 and reverse the sign).
Use eqn 3 as is.
You should get
2Mg + 2O2 + 2H2 ==> 2Mg(OH)2 which is just twice what you want. So add the delta Hs from above and take half of it.
Reverse eqn 2 and multiply by 2. (also multiply delta H by 2 and reverse the sign).
Use eqn 3 as is.
You should get
2Mg + 2O2 + 2H2 ==> 2Mg(OH)2 which is just twice what you want. So add the delta Hs from above and take half of it.
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