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Can someone help with this problem: Standard enthalpy

  1. May 9, 2006 #1
    Calculate the standard enthalpy change for the reaction

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

    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. jcsd
  3. May 9, 2006 #2
    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)

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