Enthalpy change (one question)－部落格

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Enthalpy change (one question)

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Consider the below experiment that involves the conversion of solid NaOH to a solution of common salt by 2 different routes. There are two stages in each case.Route 1:Step 1: NaOH(s) + aq ---> NaOH(aq, 4M)Step 2: NaOH(aq, 4M) + HCl(aq, 4M) ---> H2O(l) + NaCl(aq, 2M)The enthalpy change per moles of... 顯示更多 Consider the below experiment that involves the conversion of solid NaOH to a solution of common salt by 2 different routes. There are two stages in each case. Route 1: Step 1: NaOH(s) + aq ---> NaOH(aq, 4M) Step 2: NaOH(aq, 4M) + HCl(aq, 4M) ---> H2O(l) + NaCl(aq, 2M) The enthalpy change per moles of NaOH(s) used = -96.57 kJ mol^-1 Route 2: Step 1: HCl(aq, 4M) + aq ---> HCl(aq, 2M) Step 2: HCl(aq, 2M) + NaOH(s) ---> H2O(l) + NaCl(aq, 2M) The enthalpy change per moles of NaOH(s) used = -97.76 kJ mol^-1 Question: Compare the enthalpy change for the reaction NaOH(s) + HCl(aq, 4M) ---> H2O(l) + NaCl(aq, 2M) in each route and see if the enthalpy change depends on the route by which the reaction takes place.

最佳解答:

Route 1: Step 1: NaOH(s) + aq → NaOH(aq, 4M) Step 2: NaOH(aq, 4M) + HCl(aq, 4M) → H2O(l) + NaCl(aq, 2M) Add the above two equations together, and cancel the NaOH(aq, 4M) on the both sides. NaOH(s) + HCl(aq, 4M) → H2O(l) + NaCl(aq, 2M) The enthalpy change per moles of NaOH(s) used = -96.57 kJ mol^-1 Route 2: Step 1: HCl(aq, 4M) + aq → HCl(aq, 2M) Step 2: HCl(aq, 2M) + NaOH(s) → H2O(l) + NaCl(aq, 2M) Add the above two equations together, and cancel the HCl(aq, 2M) on the both sides. NaOH(s) + HCl(aq, 4M) → H2O(l) + NaCl(aq, 2M) The enthalpy change per moles of NaOH(s) used = -96.57 kJ mol^-1 In the above two different routes: NaOH(s) + HCl(aq, 4M) → H2O(l) + NaCl(aq, 2M) The enthalpy change per moles of NaOH(s) used = -96.57 kJ mol^-1 Therefore, it can be concluded that the enthalpy change does not depend on the route by which the reaction takes place.

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