# Can someone help with this problem: Standard enthalpy

1. May 9, 2006

### annayn

Calculate the standard enthalpy change for the reaction

2C8H18(l)+ 21 O2(g)-->8CO(g)+8CO2(g)+18H2O(l)

Given:
2C8H18(l)+25 O2(g)--> 16CO2(g)+18H2O(l) H=-11,020kJ/mol
2CO(g)+ O2(g)-->2CO2(g) H=-566.0kJ/mol

I just cant get it to cancel out from the given, to calculate the stardard enthalpy

2. May 9, 2006

### reincarnated_soul

In order to find standard enthalpy change u need to work thru the equations given with the enthalpy change and treat as if u r doin simultaneous equations in algebra.

U need to get the equation on the top from the two below. Since the first given equation has 2C8H18 u do not need to manipulate this one. However for the second one first multiply thru out by 4 since u need 8CO and 8CO2. So after mulitplyin thru out by 4 u get:

8CO + 4O2 -> 8CO2 H=-2264 kJ/mol

Inorder to get the equation for which u have to calculate the enthalpy change u need CO on the r.h.s of the equation. Therefore u need to reverse the equation and as such the sign of the enthalpy change for that will also change:

8CO2 -> 8CO + 4O2 H=2264 kJ/mol

Now adding this equation with the first given equation u get the required enthalpy change since adding them makes u get the equation on the top.

2C8H18 + 25O2 -> 16CO2 + 18H2O H=-11020 kJ/mol
8CO2 -> 8CO + 4O2 H=2264 kJ/mol

Resulting equation:

2C8H18 + 21O2 -> 8CO2 + 8CO + 18H2O

Therefore the enthalpy change=-11020 + 2264 kJ/mol (since u added the
=-8756 kJ/mol equations)