CHM1 12Example10 Collection

See the calculation on this YouTube video

Solution We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):

a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2a) by 2.

b) Since Eq. (1) has 1 mol H₂ on the left, we leave Eq. (2b) unchanged.

c) Since Eq. (1) has 1 mol C₂H₂ on the right, whereas there is 1 mol C₂H₂ on the left of Eq. (2c) we write Eq. (2c) in reverse.

We then have

?H_m = (-787.0 -285.8 + 1299.8) kJ

=227.0 kJ

Cancelling 5/2 O₂ on each side, the desired result is

2C(s) + H₂(g) ? C₂H₂(g)      ?H_m = 227.0 kJ