Calculate the lattice engery of sodium oxide (Na02) Solution we want Na2O so we

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we want Na2O so we can start with 2Na (s)+ 1/2O2(g) but Na needs to be in gaseous state so 1. 2Na (s)+ 1/2O2(g) => 2Na (g)+ 1/2O2(g) dH: (109*2) KJ/mol since we have 2 moles of Na next we need to break the O2 bond to create O atoms 2. 2Na (s)+ 1/2O2(g) => 2Na (g)+ O(g) dH: (499/2) KJ/mol since we have 1/2 moles of O2 next we need to ionize Na 3. 2Na (s)+ 1/2O2(g) => 2Na+ (g) + 2e- + O (g) dH: (495*2) KJ/mol since we have 2 moles of Na next O will accept the 2e- (electron affinity of O for 2e-) 4. 2Na+ (g) + 2e- + O (g) => 2Na+ + O^(2-) (g) dH: (603)KJ/mol reaction 5 is our lattice energy reaction 5. 2Na+ + O^(2-) (g) => Na2O (s) dH: unknown (to find) since we know that enthalpy is a state function, instead of doing all these steps we could do the reaction in one step and have the same enthalpy for net reaction. 2 Na (s) + O2 (g) => Na2O (s) dH: -416 kj/mol so -416 = 1 +2 +3 +4 +5 therefore, 5 = -416-(1 +2 +3 +4) viz. lattice energy lattice energy= -416-((109*2)+(499/2)+(495*2)+(603)) = -2 476.5 KJ/mol

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