Use the values of Delta H degree_r times n given for the equations Cu(s) + Cl_2

Question

Use the values of Delta H degree_r times n given for the equations Cu(s) + Cl_2 (g) rightarrow CuCl_2 (s) Delta H degree_r times n = -220.1 kJ middot mol^-1 2Cu(s) + Cl_2 (g) rightarrow 2 Cu Cl(s) Delta H degree_r times n = -137.2 kJ middot mol^-1 to calculate the value of Delta H degree_r times n for the equation Cu Cl_2 (s) + Cu(s) rightarrow 2 CuCl(s) Use the values of Delta H degree_r times n given for the equations 2Fe(s) + 3/2 O_2 (g) rightarrow Fe_2 O_3 (s) Delta H degree_r times n = -824.2 kJ middot mol^-1 3Fe(s) + 2O_2 (g) rightarrow Fe_3 O_4 (s) Delta H degree_r times n = -1118.4 kJ middot mol^-1 to calculate the value of Delta H degree_r times n for the equation 3Fe_2 O_3 (s) rightarrow 2Fe_3 O_4 (s) + 1/2 O_2 (g)

Explanation / Answer

Q.14.20: We will get the required equation by reversing the first equation and then adding the second equation.

Hence dH0rxn = - (- 220.1 kj.mol-1) + (- 137.2 kJ.mol-1)

=> dH0rxn = +82.9 kJ/mol-1 (answer)

Q.14.21: Reversing the 1st reaction and then multiplying by 3 we get

3Fe2O3(s) ------- > 6Fe(s) + 4.5 O2(g): dH0rxn = - 3x(- 824.2 kJ.mol-1) = +2472.6 kJ.mol-1 ------ (1)

Multiplying the 2nd reaction by 2 we get

6Fe(s) + 4 O2(g) -------- > 2Fe3O4(s): dH0rxn = 2x(- 1118.4 kJ.mol-1) = - 2236.8 kJ.mol-1 ------- (2)

Now we will get the desired equation by addign eqn (1) and (2)

3Fe2O3(s)  ------ > 2Fe3O4(s) + 1/2O2(g):

dH0rxn = +2472.6 kJ.mol-1 + ( - 2236.8 kJ.mol-1) = + 235.8 kJ.mol-1 (answer)

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