Calculate the enthalpy change for the reaction: Zn (s) + S (s) + 2 O2 (g) → ZnSO4 (s) ΔH = ? kJ by using the following data:
Zn (s) + S (s) → ZnS (s) ΔH = –206.0 kJ
ZnS (s) + 2 O2 (g) → ZnSO4 (s) ΔH = –776.8 kJ
My answer is -982.8 but I'm not 100% sure on that.
Tnat is correct. Here is how you do these. Add the two equations so they give you the equation you want, like this
Zn(s) + S(s) →ZnS (s) ΔH = –206.0 kJ
+ZnS(s) + 2O2(g) → ZnSO4(s) ΔH = –776.8 kJ
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Zn(s) + S(s) + 2O2(g) ==> ZnSO4(s)
Notice that ZnS(s) on the product side of equation 1 cancels with ZnS(s) on the reactant side of equation 2. I have bolded the substances that cancel to make it easier to see. By adding the two equations you get the reaction to be calculated; therefore you add the two dH values. This is a simple problem. Later you will need to reverse equations, multiply 1 or more by some number, etc.
Thank you
To calculate the enthalpy change for the given reaction, you can use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
In this case, you have two steps:
1. Zn (s) + S (s) → ZnS (s) ΔH = -206.0 kJ
2. ZnS (s) + 2 O2 (g) → ZnSO4 (s) ΔH = -776.8 kJ
To find the enthalpy change for the target reaction, you need to add the enthalpy changes of these two steps:
ΔH = -206.0 kJ + (-776.8 kJ)
ΔH = -206.0 kJ - 776.8 kJ
ΔH = -982.8 kJ
Therefore, your answer of -982.8 kJ is correct. The enthalpy change for the given reaction is -982.8 kJ.
To calculate the enthalpy change for the reaction, you can use the concept of Hess's Law, which states that the total enthalpy change for a reaction is independent of the pathway taken. In other words, you can add or subtract individual reactions to obtain the desired reaction.
In this case, you want to find ΔH for the reaction: Zn (s) + S (s) + 2 O2 (g) → ZnSO4 (s).
Given the following data:
1. Zn (s) + S (s) → ZnS (s) ΔH = –206.0 kJ
2. ZnS (s) + 2 O2 (g) → ZnSO4 (s) ΔH = –776.8 kJ
You need to reverse the first reaction because it is given in the opposite direction. To reverse it, simply change the sign of the enthalpy change:
3. ZnS (s) → Zn (s) + S (s) ΔH = +206.0 kJ
Now, you can add the equations to obtain the desired overall reaction:
4. Zn (s) + S (s) + 2 O2 (g) → ZnSO4 (s) ΔH = (3) + (2) = (206.0) + (-776.8) = -570.8 kJ
Therefore, the enthalpy change for the reaction is -570.8 kJ.
Please note that the answer you provided, -982.8 kJ, seems to be a miscalculation.