17. You have the following equations:
2S (s) + 3O2 (g) 2SO3 (g), H = 792 kJ
2S (s) + 2O2 (g) 2SO2 (g), H = ?
2SO2 (g) + O2 (g) 2SO3 (g), H = 198 kJ
What is the missing H?
To find the missing enthalpy change (H) value in the equation 2S (s) + 2O2 (g) 2SO2 (g), we can use a concept called Hess's Law. Hess's Law states that the enthalpy change for a chemical reaction is independent of the pathway from the reactants to the products and depends only on the initial and final states of the system.
In this case, we can use the given equations and manipulate them to find the missing enthalpy change.
1. First, let's reverse the equation 2SO2 (g) + O2 (g) 2SO3 (g) to get the reaction going in the opposite direction:
2SO3 (g) 2SO2 (g) + O2 (g), H = -198 kJ
2. Next, let's multiply the first equation by 2 to balance the moles of sulfur and oxygen atoms:
4S (s) + 6O2 (g) 4SO3 (g), H = 1584 kJ
3. Now, let's add these two equations together to cancel out the intermediates:
4S (s) + 6O2 (g) 4SO3 (g) + 2SO3 (g) 2SO2 (g) + O2 (g)
This will give us:
4S (s) + 6O2 (g) 2SO2 (g) + 5SO3 (g), H = 1386 kJ
4. From this combined equation, we can see that the formation of 2SO2 (g) corresponds to a net enthalpy change of 1386 kJ. However, we want to find the enthalpy change for the formation of 2SO2 (g), so we need to multiply the equation by 1/2 to get the correct stoichiometric ratio:
2S (s) + 2O2 (g) 2SO2 (g), H = 1386 kJ * (1/2) = 693 kJ
Therefore, the missing enthalpy change for the equation 2S (s) + 2O2 (g) 2SO2 (g) is 693 kJ.
To find the missing value of H in the equation 2S (s) + 2O2 (g) 2SO2 (g), we can use the concept of Hess's Law.
Hess's Law states that the enthalpy change of a reaction is the same regardless of the pathway taken.
To find the missing value, you can use the given equations to manipulate them in a way that the desired equation is formed.
Here's how to do it step-by-step:
1. Start with the equation 2S (s) + 2O2 (g) 2SO2 (g), which is the desired equation.
2. Compare it with the first equation: 2S (s) + 3O2 (g) 2SO3 (g), H = 792 kJ.
3. Notice that the reactant side is the same in both equations; however, the product side is different.
4. Multiply the first equation by 2, so that the number of oxygen molecules on the reactant side matches the desired equation.
2(2S (s) + 3O2 (g) 2SO3 (g)) 4S (s) + 6O2 (g) 4SO3 (g), H = 1584 kJ.
5. Next, consider the third given equation: 2SO2 (g) + O2 (g) 2SO3 (g), H = 198 kJ.
6. Notice that the product side of the third equation matches the product side of the equation obtained in step 4 above.
7. To cancel out the 2SO3 (g) term on the product side of the third equation, we need to flip the direction of the equation and reverse the sign of H.
2SO3 (g) 2SO2 (g) + O2 (g), H = -198 kJ.
8. Now, adding the two manipulated equations obtained from steps 4 and 7 will give you the desired equation.
4S (s) + 6O2 (g) + 2SO3 (g) 4SO3 (g) + 2SO2 (g) + O2 (g), H = 1386 kJ.
9. Simplify the equation:
4S (s) + 7O2 (g) 4SO3 (g) + 2SO2 (g), H = 1386 kJ.
10. Compare the manipulated equation with the desired equation:
4S (s) + 7O2 (g) 4SO3 (g) + 2SO2 (g), H = 1386 kJ.
2S (s) + 2O2 (g) 2SO2 (g), H = ?
11. The comparison reveals that the desired equation can be obtained by halving the manipulated equation obtained in step 9.
(4S (s) + 7O2 (g) 4SO3 (g) + 2SO2 (g))/2 (2S (s) + 2O2 (g) 2SO2 (g)), H = 693 kJ.
Therefore, the missing value of H in the equation 2S (s) + 2O2 (g) 2SO2 (g) is 693 kJ.