Please help..i dont get this enthalpy question?

Question

determine the enthalpy of combustion of 1-propanol, C3H7OH(l) using the following data

a) C3H8 + 5O2 = 3CO2 + 4H2O -2219.1 Kj/mol C3H8

b) C(graphite) + O2 = CO2 -393.5 kj/mol C

c) H2 + 1/2O2 = H2O -285.8 Kj/mol H2

d) C3H6 + H2O = C3H7OH -36.8 Kj/mol C3H6

e) C3H6 + H2 = C3H8 -125 kj/mol C3H6

please explain and please help..

i am having trouble with enthalpy..please help

thanks

Answer 1

Enthalpy of reaction is just the amount of heat given off (or absorbed) by that reaction. What they want to know is how much heat is given off (or absorbed) when C3H7OH(l) is combusted. First we need to write and balance the combustion reaction. The combustion products of any organic molecule containing only C and H (or O) are CO2(g) and H2O(g).

So:.......C3H7OH(l) + O2 =>  CO2(g) + H2O(g)

Obviously that's not a balanced equation.

We have 3 C on the left, so let's put a 3 in front of CO2 on the right.

.....C3H7OH(l) + O2(g) =>  3CO2(g) + H2O(g)

We have 8 H on the left; to get 8 H on the right, multiply H2O by 4:

C3H7OH(l) + O2(g) =>  3CO2(g) + 4H2O(g)

We have 10 O on the right; to get 10 on the left, we need to multiply O2 by 4.5:

C3H7OH(l) + 4.5O2(g) => 3CO2(g) + 4H2O(g)

The equation is now balanced.

We need to rearrange the given equations (a) through (e) to end up with what we have above.

Look at equation (d); it has C3H7OH on the right. Let's reverse it (and its delta H sign) to get C3H7OH on the left like it is in our combustion reaction. (delta = ^).

(d)-1 : C3H7OH => C3H6 + H2O  ^H =  +36.8 kJ

Notice that equation (a) has both CO2 and H2O on the right side, just the way we want it.

(a) C3H8 + 5O2 => 3CO2 + H2O ^H = -2219.1 kJ

When we reversed equation (d) to make (d)-1, we put a C3H6 on the right that isn't in our combustion equation. We need to add equation (e) which has a C3H6 on the left to cancel it out.

(e) C3H6 + H2 => C3H8  ^H = -125 kJ

Notice that equation (e) puts a C3H8 on the right that cancels the one on the left in (a).

Let's add (d)-1, (a), and (e) together to see what we have:

C3H7OH + C3H8 +5O2 + C3H6 + H2 =>

C3H6 + H2O+ 3CO2 + 4H2O + C3H8

C3H7OH + 5O2 + H2 => 5H2O + 3CO2

^H = (36.8 - 2219.1`- 125) = -2307.3 kJ

This matches our combustion equation except that there is an extra H2 and 1/2 O2 on the left and an extra H2O on the right. If we SUBTRACT equation (c), that will do it.

-(c): -(H2 + 1/2 O2 => H2O)    ^H = 285.8 kJ

Adding (d)-1, (a), (e), and -(c) gives our final result. ^H = -2307.3 kJ + 285.8 = -2021.5 kJ

per mole of C3H7OH combusted. Since ^H is negative, the reaction GIVES OFF heat (is exothermic).