Enter your answer in the provided box. from the following data, c(graphite) + o2(g) → co2(g)δh o rxn = −393.5 kj/mol h2(g) + 1 2 o2(g) → h2o(l)δh o rxn = −285.8 kj/mol 2c2h6(g) + 7o2(g) → 4co2(g) + 6h2o(l)δh o rxn = −3119.6 kj/mol calculate the enthalpy change for the reaction below: 2 c(graphite) + 3h2(g) → c2h6(g)

final reaction to get is 2C(graphite) + H2(g) → C2H6(g)

First step: 2C(graphite) + 2O2(g) → 2CO2 (g)   ΔH = (-393.5 x 2)

3H2(g) + 3/2O2(g) → 3H2O(l)  ΔH = (-285.8 x 3)

2CO2(g) + 3H2O(l) → C2H6(g) + 7/2O2(g)  ΔH = (-3119.6 X -1/2)

Second step: 2C(graphite) + 2O2(g) + 3H2(g) + 3/2O2 (g) + 2CO2(g) + 3H2O(l) → 2CO2(g) + 3H2O(l) + C2H6(g) + 7/2O2(g)

Third step: 2C(graphite) + O2(g) → C2H6(g)

ΔH final = (-393.5 x 2) + (-285.8 x 3) + (-3119.6 x -1/2) = -84.6 KJ/mol

Explanation: Using Hess Law

In the first step, rearrange the 3 given reactions in a way that they match the final reaction by multiplying or dividing (also the ΔH) by a given number and by reverse (if so the multiply by -1).

In the second step, add all 3 reactions with reactant on their side and product on theirs.

In 3rd step, cancel out similarity in the product and reactant(such as there are (2+3/2 = 7/2)O2 which cancels with 7/2O2 in the product) .

finally, add all 3 ΔH to find the final one which is -84.6 KJ/mol