Question

Given the following reactions and subsequent delta H values,

C2H6 (g) -> C2H2 (g) + 2H2 (g) delta H= 283.5 kJ

H2 (g) + 1/2 O2 (g) -> H2O (g) delta H= -213.7 kJ

2CO2 (g) + 3H2O (g) -> C2H6 (g) + H2O (g) delta H= 849 kJ

Find the delta H for the reaction: C2H2 (g) + 5/2 O2 (g) -> 2CO2 (g) + H2O (g)

Answer 1

See Here we need to form the final equation C2H2 (g) + 5/2 O2 (g) -> 2CO2 (g) + H2O (g)....(4)

for this we need to caluclate the delta H . We will make use of Hess's law .

First step is to rearrange all the above reactions such that on adding them we will end up with the formation of our final equation as follows:

Number the reactions as (1) (2) (3)

C2H6 (g) -> C2H2 (g) + 2H2 (g) delta H= 283.5 kJ....(1)

H2 (g) + 1/2 O2 (g) -> H2O (g) delta H= -213.7 kJ.....(2)

2CO2 (g) + 3H2O (g) -> C2H6 (g) + H2O (g) delta H= 849 kJ.......(3)

Reverse the reaction (1) & (3) tp (a) & (c) respectively as follows:

C2H2 (g) + 2H2 (g) -> C2H6 (g) delta H = - 283.5 kJ......(a)

C2H6 (g) + 7/2 O2 (g) -> 2CO2 (g) + 3H2O (g) delta H = - 849 kJ......(c) [ 7/2 O2 I have added , as this equation given in the question s incorrect ]

Now multiply the the equation (2) by 2 and write it as (b) as follows :

2H2 (g) + O2 (g) -> 2H2O (g) delta H = - 427.4 kJ..........(b)

Now add equations (a) + (c) and subtract (b) from the sum of (a) and (c) we will get the equation final equation (4) , and the delta H for that is as follows:

delta H for equation (4) = delta H of equation [(a) + (c) - (b)]

delta H = - 283.5 kJ - 849 kJ - (-427.4)

delta H = -705.1 KJ