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Whats the standard reaction enthalpy for the partial combustion of methane to carbon...
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The first step in any problem with a chemical reaction is to make sure it's balanced, which this reaction is. To find the enthalpy of a reaction, we need to know the enthalpies of formation for all species in the reaction. 2CH4(g) + 3O2(g) −> 2CO(g) + 4H2O(l)
Once we know the enthalpies of formation, we can set up our calculation by taking the enthalpies of formation of the product and subtracting the enthalpies of formation of the reaction. Remember that the enthalpy of formation of elements in their standard state is always zero. We also need to include the coefficients by multiplying each enthalpy of formation by the substance's coefficient from the balanced equation.
To keep things less cluttered, I'll use a capital D to represent the delta symbol (triangle).
DH(rxn) = 2 DH(CO(g)) + 4 DH(H2O(l)) - 2 DH(CH4(g)) - 3 DH(O2(g))
Now, we need to find the enthalpies of formation for each substance. Note that the phase does matter. Liquid water and gas phase water will have different enthalpies of formation. Since the enthalpy of formation of water isn't given in the problem, we'll need to check another source for that information (see link below).
DH(rxn) = 2(-283 kJ/mo) + 4(-285 kJ/mol) - 2(-890 kJ/mol) - 3(0)
Watch your signs carefully as you work out the math!
DH(rxn) = -566 - 1140 + 1780 - 0
DH(rxn) = 74 kJ/mol
Now we know that the standard enthalpy for the reaction given was 74 kJ/mol.
Posted by mlsiasebs on February 25, 2012 at 10:21 PM (Answer #1)
That's not one of the choices I am afraid.
Posted by abixo on March 8, 2012 at 1:18 AM (Answer #2)
Methane is a chemical compound with the chemical formula CH4. It is the simplest alkane, the principal component of natural gas, and probably the most abundant organic compound on earth. The relative abundance of methane makes it an attractive fuel. However, because it is a gas at normal conditions, methane is difficult to transport from its source.
Posted by young2000 on February 25, 2012 at 5:35 PM (Answer #3)
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