During volcanic eruptions, hydrogen sulfide gas is released and is oxidized by the oxygen in the air yielding sulfur dioxide gas and water vapour. Calculate the standard enthalpy change for this reaction given the following:
3S(rhombic) + 2H2O(g) → 2H2S(g) + SO2(g)
ΔH° = 146.9 kJ
S(rhombic) + O2(g) → SO2(g)
ΔH° = –296.4 k
To determine the standard enthalpy change for the reaction, we can use the Hess's Law. Hess's Law states that the overall enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.
The given reaction can be broken down into two steps:
Step 1: S(rhombic) + O2(g) → SO2(g) (ΔH° = –296.4 kJ)
Step 2: 2H2O(g) + SO2(g) → 2H2S(g) (ΔH° = 146.9 kJ)
To find the enthalpy change for the overall reaction, we need to sum the enthalpy changes of both steps.
Step 1 has a negative enthalpy change, so we multiply it by -1 to reverse its sign:
-1 × ΔH° (step 1) = -1 × (-296.4 kJ) = +296.4 kJ
Now we can sum the enthalpy changes:
ΔH° (overall) = ΔH° (step 1) + ΔH° (step 2)
ΔH° (overall) = 296.4 kJ + 146.9 kJ
ΔH° (overall) = 443.3 kJ
Therefore, the standard enthalpy change for the given reaction is 443.3 kJ.
To calculate the standard enthalpy change for the given reaction, we need to consider the enthalpies of the individual reactions and use them to determine the overall enthalpy change.
First, let's assign a value to the enthalpy change for the reaction of S(rhombic) with oxygen to form SO2:
ΔH1 = -296.4 kJ
Next, we consider the reaction of 3 moles of S(rhombic) reacting with 2 moles of water to produce 2 moles of H2S and 1 mole of SO2. The enthalpy change for this reaction is given as:
ΔH2 = 146.9 kJ
Based on stoichiometry, we can see that the given reaction involves the reaction of 2 moles of S(rhombic), so we need to multiply ΔH2 by a factor of 2:
2 * ΔH2 = 2 * 146.9 kJ = 293.8 kJ
Now, we can add up the enthalpy changes of the two reactions:
ΔH_total = ΔH1 + 2 * ΔH2
ΔH_total = -296.4 kJ + 293.8 kJ
ΔH_total = -2.6 kJ
Therefore, the standard enthalpy change for the given reaction is -2.6 kJ.
Reverse eqn 1 and add to 3x eqn 2. That gives you the equation you want. Then for dH, change the sign of eqn 1 and add to 3x dH for eqn 2.
By the way, you can't get what you want UNLESS you balance that equation you want.
2H2S + 3O2 ==> 2SO2 + 2H2O