Determine the enthalpy for the partial combustion of methane to carbon monoxide-
2CH4(g) + 3O2(g) = 2CO(g) + 4H2O(l) from
change in enthalpy (CH4,g) = -890. kJ mol-1 and
change in enthalpy (CO,g) = -283.0 kJ mol-1
delta H rxn = (delta H products-delta H reactants). You will need to look up the delta H for H2O unless you have it and didn't post it.
To determine the enthalpy for the partial combustion of methane to carbon monoxide, you need to use the enthalpy values given for methane (CH4) and carbon monoxide (CO), as well as the balanced chemical equation.
The chemical equation for the reaction is:
2CH4(g) + 3O2(g) = 2CO(g) + 4H2O(l)
Given:
Change in enthalpy (∆H) for methane (CH4): -890 kJ mol-1
Change in enthalpy (∆H) for carbon monoxide (CO): -283 kJ mol-1
Step 1: Calculate the enthalpy change for the given reaction using the enthalpy values provided.
The enthalpy change (∆H) for the reaction can be calculated as follows:
∆H = (2 * ∆H(CH4)) + (2 * ∆H(H2O)) - (2 * ∆H(CO)) - (3 * ∆H(O2))
Since there are 4 moles of water formed in the reaction, we multiply the enthalpy change of water by 2.
∆H = (2 * -890 kJ mol-1) + (4 * -285.8 kJ mol-1) - (2 * -283 kJ mol-1) - (3 * 0 kJ mol-1)
∆H = -1780 kJ + (-1143.2 kJ) + 566 kJ
∆H = -1780 kJ - 1143.2 kJ + 566 kJ
∆H = -2357.2 kJ
Therefore, the enthalpy change (∆H) for the partial combustion of methane to carbon monoxide is -2357.2 kJ.
To determine the enthalpy for the partial combustion of methane to carbon monoxide, you need to calculate the overall change in enthalpy using the given enthalpies of formation.
The enthalpy change for a reaction can be determined by subtracting the sum of the enthalpies of the reactants from the sum of the enthalpies of the products.
First, let's calculate the enthalpy change of the reactants:
2CH4(g): Since the given enthalpy change is for one mole of methane, you need to multiply it by the stoichiometric coefficient of methane in the reaction equation:
Enthalpy change for 2 moles of methane = -890 kJ/mol × 2 = -1780 kJ
3O2(g): The enthalpy change for molecular oxygen (O2) is not given, but it is 0 kJ/mol because it is the most stable form of oxygen at standard conditions.
Next, calculate the enthalpy change of the products:
2CO(g): Multiply the given enthalpy change for carbon monoxide (CO) by the stoichiometric coefficient:
Enthalpy change for 2 moles of carbon monoxide = -283 kJ/mol × 2 = -566 kJ
4H2O(l): The enthalpy change for water (H2O) is not known, but you can look it up in tables. The enthalpy change for the liquid state is commonly used. The enthalpy change of formation for liquid water is -285.8 kJ/mol.
Now, sum up the enthalpies of the products and subtract the enthalpies of the reactants to get the overall enthalpy change:
(-566 kJ) + (-285.8 kJ × 4) = -566 kJ - 1143.2 kJ = -1709.2 kJ
Therefore, the enthalpy change for the partial combustion of methane to carbon monoxide is -1709.2 kJ.