1) A(s)+O2(g) →Ao2; ΔH^0=-394kj/mol

2)AO(g)+ $\frac{1}{2}$O2(g) →AO2; ΔH^0=-283kj/mol

3)A(s)+ $\frac{1}{2}$O2(g) →AO(g); ΔH^0=-Xkj/mol

Find the value of X? I want to know How to find this value.


closed as off-topic by airhuff, Martin - マーチン May 16 '17 at 6:11

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You need to reverse the 2nd reaction. This leaves you with Ao2 ---> frac{1}{2}O2(g) + AO(g). And since you're reversing the reaction, you have to switch the sign of delta H, so it becomes 283 kj/mol (endothermic). Once you do this, you can cancel out the AO2 and frac{1}{2}O2 with the 1st reaction. This leaves you with your desired reaction, that is, A(s) + frac{1}{2}O2 ---> AO(g). That is the reaction you want. To find delta H for that reaction, simply add the delta H's from the previous reactions. So, -394+283 = -111 kj/mol. So your X-value is 111. This process works because of Hess's Law. In fact, it is a classic example of how Hess's law would be implemented in chemistry.

To wrap up: The reason we reversed the 2nd reaction was because doing that allows us to cancel out some stuff with the 1st reaction in order to get the 3rd reaction. Our goal is to get the third reaction you described. The reason we switched the sign of delta H (from -283 to 283) is because if a reaction is exothermic (negative delta H) in one direction, it is endothermic (positive delta H) in the reverse direction. That is just a rule of thumb; you can obtain further understanding of such concepts by learning about bond energies, activation energies and the like.

You can get further understanding of why Hess's Law works, as well as practice, here.


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