Given the following data:
S(s) + 3/2 O2(g)-->SO3(g) ∆H = -395.2 kJ
2SO2(g) + O2(g) -->2SO3(g)∆H = -198.2 kJ
Calculate ∆H for the reaction
S(s) + O2(g) --> SO2(g)
Could someone please tell me how you would go about doing this type of problem?
To calculate the enthalpy change (∆H) for the reaction:
S(s) + O2(g) --> SO2(g), we can use Hess's law. Hess's law states that the enthalpy change of a reaction is independent of the pathway taken. In other words, if we can express the desired reaction as a combination of given reactions with known enthalpy changes, we can manipulate and combine these reactions to obtain the overall enthalpy change.
To solve this problem, we need to find combinations of the given reactions that cancel out the intermediate species, which in this case is SO3(g).
First, let's multiply the second reaction by 2 to cancel out SO3(g):
2SO2(g) + 2O2(g) --> 4SO3(g) (∆H = -2(198.2 kJ) = -396.4 kJ)
Now, note that in the first reaction, 1 mole of SO3(g) is formed, while in the modified second reaction it's 4 moles of SO3(g). By canceling out the intermediate, we can obtain the overall equation:
4(S(s) + O2(g) --> SO2(g)) + 2SO3(g) --> 4SO3(g) (∆H = 4∆H1 + ∆H2)
Since we want the reaction S(s) + O2(g) --> SO2(g), we can see that the coefficient of this reaction in the overall equation is 4. Hence:
∆H = 4(-395.2 kJ) + (-396.4 kJ)
∆H = -1580.8 kJ - 396.4 kJ
∆H = -1977.2 kJ
Therefore, the enthalpy change (∆H) for the reaction S(s) + O2(g) --> SO2(g) is -1977.2 kJ.
To calculate the ΔH for the reaction:
S(s) + O2(g) --> SO2(g)
using the given data, we can use Hess's Law. Hess's Law states that the sum of the enthalpy changes for a series of reactions is equal to the enthalpy change of the overall reaction.
Step 1: Write the given equations:
1. S(s) + 3/2 O2(g) --> SO3(g) ΔH = -395.2 kJ
2. 2SO2(g) + O2(g) --> 2SO3(g) ΔH = -198.2 kJ
Step 2: Manipulate the equations to cancel out the desired compound (SO2 in this case):
- Multiply the second equation by 2 so that the number of moles of SO2 matches the desired equation.
- Reverse the first equation to match the desired equation.
The modified equations become:
1. -SO3(g) --> S(s) + 3/2 O2(g) ΔH = +395.2 kJ
2. 2SO3(g) --> 2SO2(g) + O2(g) ΔH = +198.2 kJ
Step 3: Sum up the modified equations:
-SO3(g) + 2SO3(g) --> S(s) + 3/2 O2(g) + 2SO2(g) + O2(g)
Step 4: Simplify the equation by canceling out the terms on both sides:
-SO3(g) + 2SO3(g) --> S(s) + 5/2 O2(g) + 2SO2(g)
SO3(g) - SO3(g) --> S(s) + 5/2 O2(g) + 2SO2(g)
0 --> S(s) + 5/2 O2(g) + 2SO2(g)
Step 5: Determine the ΔH for the overall reaction:
Since an enthalpy change of 0 kJ corresponds to no change in energy, the ΔH for the overall reaction is 0 kJ.
Therefore, ΔH for the reaction: S(s) + O2(g) --> SO2(g) is 0 kJ.
Multiply eqn 1 by 2. Add the reverse of eqn 2. This give you twice the equation you want so divide everything by 2.
For dH. dH 1 + (-dH 2) = 2*dH total and divide by 2.