Can someone help with this problem: Standard enthelpy

[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 can't get it to cancel out from the given, to calculate the stardard enthalpy

 

[reincarnated_soul]

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 can't get it to cancel out from the given, to calculate the stardard enthalpy

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 doing 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)

 

[Blueshift5]

change for the reaction.

To calculate the standard enthalpy change for this reaction, we can use the following equation:

ΔH°(reaction) = ΣnΔH°(products) - ΣmΔH°(reactants)

Where ΔH° is the standard enthalpy change, n is the number of moles of products, and m is the number of moles of reactants.

First, let's balance the given reaction to match the reaction in the problem statement:

2C8H18(l)+ 25 O2(g)--> 16CO2(g)+18H2O(l)

Next, we can use the given values for ΔH° to calculate the standard enthalpy change for each individual reaction:

ΔH°(reaction 1) = 2ΔH°(2C8H18(l)) + 25ΔH°(O2(g)) - 16ΔH°(CO2(g)) - 18ΔH°(H2O(l))
ΔH°(reaction 1) = 2(0 kJ/mol) + 25(-11,020 kJ/mol) - 16(-566.0 kJ/mol) - 18(0 kJ/mol)
ΔH°(reaction 1) = -275,500 kJ/mol

Now, we can use this value to calculate the standard enthalpy change for the reaction in the problem statement:

ΔH°(reaction 2) = 2ΔH°(2C8H18(l)) + 21ΔH°(O2(g)) - 8ΔH°(CO(g)) - 8ΔH°(CO2(g)) - 18ΔH°(H2O(l))
ΔH°(reaction 2) = 2(0 kJ/mol) + 21(-11,020 kJ/mol) - 8(-566.0 kJ/mol) - 8(0 kJ/mol) - 18(0 kJ/mol)
ΔH°(reaction 2) = -231,300 kJ/mol

Therefore, the standard enthalpy change for the reaction is -231,300 kJ/mol. It is important to note that the values given in the problem statement may not be accurate, as they do not match the