MnO2(s) -> MnO(s) + 1/2O2(g) Delta H1 = +32.5 kcal
MnO2(s) + Mn(s) -> 2MnO(s) Delta H2 = -59.0 kcal
Calculate Delta H final of MnO2 in KJ.
The answer is not as important to me as the explanation. So, can someone please show me how to do this type of a problem?
To calculate the overall enthalpy change (ΔH) for the reaction, you need to apply the Hess's Law of Heat Summation. Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken.
In this case, you're given two reactions along with their respective enthalpy changes:
1. MnO2(s) → MnO(s) + 1/2O2(g) ΔH1 = +32.5 kcal
2. MnO2(s) + Mn(s) → 2MnO(s) ΔH2 = -59.0 kcal
To calculate ΔHfinal for the reaction MnO2(s) → 2MnO(s), you need to manipulate the given reactions and their enthalpy changes to cancel out the common compounds and obtain the desired reaction.
Step 1: Reverse equation 1 to match the desired equation:
MnO2(s) + 1/2O2(g) → MnO(s) ΔH1 = -32.5 kcal (reversed the sign)
Step 2: Multiply equation 2 by 2 to match the desired equation:
2[MnO2(s) + Mn(s) → 2MnO(s)] ΔH2 = -118.0 kcal (multiplied ΔH2 by 2)
Now you have the following equations:
1. MnO2(s) + 1/2O2(g) → MnO(s) ΔH1 = -32.5 kcal
2. 2[MnO2(s) + Mn(s) → 2MnO(s)] ΔH2 = -118.0 kcal
Step 3: Add the two equations to obtain the desired equation:
MnO2(s) + 1/2O2(g) + 2[MnO2(s) + Mn(s)] → MnO(s) + 2MnO(s)
Now you can add the enthalpy changes together:
ΔHfinal = ΔH1 + ΔH2
ΔHfinal = -32.5 kcal + (-118.0 kcal)
ΔHfinal = -150.5 kcal
To convert ΔHfinal from kcal to kJ, divide by 1,000:
ΔHfinal = -150.5 kcal / 1,000
ΔHfinal = -0.1505 kJ
Therefore, the final enthalpy change for the reaction MnO2(s) → 2MnO(s) is -0.1505 kJ.