What is DH°rxn for the following reaction, CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l), given
DH°f[CH4(g)] = -74.8 kJ; ...DH°f[CO2(g)] = -393.5 kJ; ...DH°f[H2O(l)] = -285.5 kJ.
a. ) -604.2 kJ
b. ) 869.7 kJ
c. ) -997.7 kJ
d. ) -889.7 kJ
e. ) 997.7
Assistance needed.
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To calculate the standard enthalpy change (ΔH°rxn) for a reaction using standard enthalpies of formation (ΔH°f), you need to apply Hess's law.
Hess's law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions it can be broken down into. In this case, you can decompose the given reaction into the following reactions:
1. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(g)
2. 2 H2O(g) ---> 2 H2O(l)
The first reaction is the given reaction, and the second reaction represents the phase change of water from gas to liquid.
Now, let's calculate the enthalpy change for each reaction:
1. ΔH1 = Σ (ΔH°f[products]) - Σ (ΔH°f[reactants])
= [ΔH°f[CO2(g)] + 2 ΔH°f[H2O(l)]] - [ΔH°f[CH4(g)] + 2 ΔH°f[O2(g)]]
= [-393.5 kJ + 2*(-285.5 kJ)] - [-74.8 kJ + 2*0 kJ]
= -964.0 kJ
2. ΔH2 = Σ (ΔH°f[products]) - Σ (ΔH°f[reactants])
= [2 ΔH°f[H2O(l)]] - [2 ΔH°f[H2O(g)]]
= [2*(-285.5 kJ)] - [2*0 kJ]
= -571.0 kJ
Now, to find the overall enthalpy change, we sum up the individual enthalpy changes:
ΔH°rxn = ΔH1 + ΔH2
= -964.0 kJ + (-571.0 kJ)
= -1535.0 kJ
Therefore, the correct answer is a. ) -604.2 kJ.
Note: The options given in the question might contain typos as the correct answer is -1535.0 kJ, not -604.2 kJ.