From the following data, c(graphite) + o2(g) ⟶co2(g) δh°rxn = −393.5 kj/mol h2(g) + 12 o2(g) ⟶h2o(l) δh°rxn = −285.8 kj/mol 2c2h6(g) + 7o2(g) ⟶4co2(g) + 6h2o(l) δh°rxn = −3119.6 kj/mol calculate the enthalpy change for the reaction 2c(graphite) + 3h2(g) ⟶c2h6(g

Explanation:

Hi!

Ur Data

C(graphite) + O2(g) --> CO2(g) ΔH = -393.5 kJ

H2(g) + 0.5 O2(g) --> H2O(l) ΔH = -285.8 kJ

2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(l) ΔH = -3119.6 kJ

calculate the enthalpy change for the reaction

2 C(graphite) + 3 H2(g) --> C2H6(g) ΔH = ?

take

2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(l)

and inverse it:  

4 CO2(g) + 6 H2O(l) ---> 2 C2H6(g) + 7 O2(g)  when you inverse it you change the AH aswell to a + sign. ΔH =  + 3119.6 kJ

Multiply with 4 the reaction:

C(graphite) + O2(g) --> CO2(g) ΔH = -393.5 kJ

that would give you:  

4C(graphite) + 4O2(g) --> 4CO2(g) ΔH = -393.5 kJ * 4 = -1576 kj

then multply with 6 :  H2(g) + 0.5 O2(g) --> H2O(l) ΔH = -285.8 kJ

that would give you:

6H2(g) + 3 O2(g) --> 6H2O(l)  ΔH = -285.8 kJ * 6    -->  ΔH = -1710 kJ

ur reactions are:

4C(graphite) + 4O2(g) --> 4CO2(g) ΔH = -393.5 kJ * 4 = -1576 kj

6H2(g) + 3 O2(g) --> 6H2O(l)  ΔH = -285.8 kJ * 6    -->  ΔH = -1710 kJ

4 CO2(g) + 6 H2O(l) ---> 2 C2H6(g) +   7 O2(g) ΔH =  3119.6 kJ

4C + 6H2 > 2 C2H7   AH= -1872 kj     but you want 2 C(graphite) + 3 H2(g) --> C2H6(g) so multiply it with 1/2 and ur reaction is:

2C + 3H2 > C2H7   AH= -1872 kj * 1/2 = -936 kj

Answer= -936 kj

Hope this helps you a bit

the increase of protons   hope this