Enthalpy change (one question)

知識就是力量

2011-03-11 1:36 am

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.

回答 (1)

pingshek

2011-03-11 6:37 am

✔ 最佳答案

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.

參考: miraco

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