J.R. S. answered • 10/26/19
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We need more information. Do you have a table of values for standard enthalpy of formation. Did they give you values that you neglected to include.
J.R. S.
tutor
(-411.2 + -285.8) - (92.3 + 425.8) =
-697 - (-518) = -179 kJ. It should be sum of products minus sum of reactants.
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10/26/19
Kinsui U.
Yes, I did forget the table. My apologies. ∆Hf° kJ/mol HCL(g) = -92.3 ∆Gf° kJ/mol HCl(g) = -95.3 S° J/(mol·K) HCl (g) = 186.9 ∆Hf° kJ/mol NaOH(s) = –425.8 ∆Gf° kJ/mol NaOH(s) = –379.7 S° J/(mol·K) NaOH(s) = 64.4 ∆Hf° kJ/mol NaCL(s) = –411.2 ∆Gf° kJ/mol NaCL(s) = –384.1 S° J/(mol·K) NaCL(s) = 72.1 ∆Hf° kJ/mol H2O(l) = –285.8 ∆Gf° kJ/mol H2O(l) –237.1 S° J/(mol·K) H2O(l) = 70.0 Sorry again.10/26/19