<p>how do you calculate enthalpy change for combustion of methane?</p>
<p>C-C is 83 Kcal/mole
C-H is 99 Kcal/mole
C=O is 174 Kcal/mole
O-H is 111 Kcal/mole
O=O is 118 Kcal/mole
C-O is 84 Kcal/mole</p>
<p>thanks!</p>
<p>Delta H=(Sum of delta H of all bonds broken)-(Sum of delta H of all bonds created)</p>
<p>So, write out your equation for combustion of methane, find the bonds broken and bonds created and use the given values to find delta H of the total reaction.</p>
<p>k thanks.</p>
<p>If I have to find the delta H of ethane and I solve the answer for 2C2H6 + …->…, do I have to divide my answer by 2 too or no?</p>
<p>Is this the same type of problem? If it is, that’s not exactly how you solve it.</p>
<p>Let me explain the combustion of methane one (just in case):</p>
<p>Bascially, you know that combustion of a hydrocarbon has these reactants and products:</p>
<p><strong>(Hydrocarbon)+</strong>O2–><strong>CO2+</strong>H2O</p>
<p>Well, your hydrocarbon is CH3, methane:</p>
<p><strong>CH3+</strong>O2–><strong>CO2+</strong>H2O</p>
<p>Balance:</p>
<p>4CH3+7O2–>4CO2+6H2O</p>
<p>Now, let’s figure out the bonds in each molecule:</p>
<p>In CH3, you have 3 C-H bonds.
In O2, you have one O=O bond.
In CO2, you have 2 C=O bonds.
In H2O, you have 2 H-O bonds.</p>
<p>Multiply the number of bonds per molecule by the coefficients. Then, calculate the bond energies of each molecule and subtract product energies by reactant energies.</p>
<p>If that explanation was completely unnecessary and your ethane question was completely unrelated, then yes. To find delta H(form) for ethane, you need to divide by 2.</p>
<p>methane is CH4</p>
<p>and i didn’t know about this equation</p>
<p><strong>(Hydrocarbon)+</strong>O2–><strong>CO2+</strong>H2O</p>
<p>thanks a lot!</p>
<p>Oops, I accidentally put (and balanced) for CH3.</p>
<p>But the combustion of any hydrocarbon is the hydrocarbon plus O2 which always makes carbon dioxide and water.</p>