If the ÄH of reaction for the following 2ZnS (s) + O 2(g) = 2ZnO (s) + 2S (s) is -290 kJ, then what is the
ÄH of reaction for the following reaction: ZnO (s) + S (s) = ZnS (s) + 1/2O 2(g)?
Would i flip the first equation? then after that i am lost
The question equation is just the reverse of the original equation listed except it's 1/2 of that. So you take 1/2 of the original delta H, then take the negative of the result.
The original is -290 kJ so it is just 1/2 x +290 kJ.
To find the ΔH of the second reaction: ZnO (s) + S (s) = ZnS (s) + 1/2O2 (g), we can use the concept of Hess's Law.
Hess's Law states that if a reaction can be expressed as a combination of other reactions, the overall enthalpy change of the reaction will be the sum of the enthalpy changes of the individual reactions.
In order to apply Hess's Law, we need to manipulate the given equation to match the format of the first equation.
First, we can reverse the first equation by flipping it, making the reactants products and the products reactants.
2ZnO (s) + 2S (s) = 2ZnS (s) + O2 (g)
Now, we need to multiply the second equation by a factor in order for the number of moles of the compounds involved to match the coefficients in the first equation.
ZnO (s) + S (s) = ZnS (s) + 1/2O2 (g) --> Multiply by 2
2ZnO (s) + 2S (s) = 2ZnS (s) + O2 (g)
Now, we can see that both equations have the same products and reactants.
Since we reversed the first equation and multiplied the second equation, we also need to reverse the sign of the ΔH for the first equation to match the second equation.
Therefore, the ΔH of the second reaction is the negative of the ΔH of the first reaction.
So, the ΔH of the second reaction is -(-290 kJ), which simplifies to +290 kJ.