Compare your experimental value with the one you have just calculated. The corre

Question

Compare your experimental value with the one you have just calculated. The correct value is only -3.9 kJ/mol. Try to explain any discrepancy between the experimental and calculated values and between these values and the correct value. Calculate the enthalpy change to be expected for the reaction NaCl(s) rightarrow NaCl(aq) where (s) and (aq) mean solid and aqueous, respectively. Use Hess's law, one of the three enthalpy changes that was measured in this experiment, and the data from the following table

Explanation / Answer

Answer – Given, reaction – NaCl(s) ----> NaCl(aq)

We need to use the given data from the table and Hess’s law to calculate the enthalpy of this reaction. We need to added the equation in such way that the overall reaction must be remaining the above one.

First we need to reaction number 2 and 6

Na(s) + ½ O2(g) + ½ H2(g) ----> NaOH(s)   H = -426.8 kJ/mol

                          NaOH(s) -----> NaOH(aq) H = -41.8 kJ/mol

Na(s) + ½ O2(g) + ½ H2(g) ----> NaOH (aq) H = -468.6 kJ/mol …..A

Now we need to added reaction 1 and 5

½ H2(g) + ½ Cl2(g) ------> HCl(g)   H = -92.3 kJ/mol

                HCl(g) -------> HCl(aq)   H = -75.2 kJ/mol

½ H2(g) + ½ Cl2(g) ------> HCl(aq)   H = -167.5 kJ/mol …….B

Now we need to reverse the reaction 3

NaCl(s) -----> Na(s) + ½ Cl2(g)   H = 411.1 kJ/mol …..C

Now we need to add A and B

Na(s) + ½ O2(g) + ½ H2(g) ----> NaOH(aq) H = -468.6 kJ/mol …..A

½ H2(g) + ½ Cl2(g) ------> HCl(aq)   H = -167.5 kJ/mol …….B

Na(s) + ½ O2(g) + H2(g) + ½ Cl2(g) ---> NaCl(aq) + H2O(l) H = -636.1kJ/mol..D

Reverse the reaction 4

H2O(l) -----> H2(g) + ½ O2(g)   H = 285.8 .....E   

Now add C, D and E

                                       NaCl(s) -----> Na(s) + ½ Cl2(g)   H = 411.1 kJ/mol …..C

Na(s) + ½ O2(g) + H2(g) + ½ Cl2(g) ----> NaCl(aq) + H2O(l) H = -636.1kJ/mol....D

                                          H2O(l) -----> H2(g) + ½ O2(g) H = 285.8 ....E

NaCl(s) -----> NaCl(aq) H = 60.8 kJ/mol

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