## anonymous one year ago Using the table below, what is the change in enthalpy for the following reaction when all reactants and products are in the gaseous state? Carbon Monoxide + Oxygen --> Carbon Dioxide

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1. anonymous

Substance ΔHf (kJ/mol) CO2 (g) -393.5 CO (g) -110.5 O2 (g) 0.0 O3 (g) 142.0 H2O (g) -241.8 H2O (l) -285.8 CH4 (g) -74.86

2. Photon336

We balance our equation first: We notice that we have 1 carbon atom on both sides of the equation but we have 3 atoms of oxygen on the reactant side and 2 atoms of oxygen on the product side. what we can do is multiply the O2 by the coefficient 1/2 so that we have two atoms of oxygen on either side. $CO(g) + \frac{ 1 }{ 2 } O _{2}(g) \rightarrow CO _{2} (g)$

3. Photon336

but you can also consider that most of the time fractions aren't in your coefficients so another way is to multiply both the carbon monoxide CO and CO2 by 2. to get the following. $2CO(g) + O _{2}(g) \rightarrow 2CO _{2}(g)$ $\sum_{r}^{?} products \Delta H - \sum_{?}^{?} reactants \Delta H.$ that formula above means you take the sum of the enahalpy from the products and subtract it from the reactants. you have 2 moles of CO2 and 2 moles of CO so how would you take that into account and get the answer?