In this example question, it says if the formation equation for boron trioxide is given as the following: 4 moles B solid + 3 moles O2 gas → 2 moles boron trioxidesolid has an enthalpy of - 2547 kilojoules, what will be the new enthalpy value when it's rearranged? All right. So we have to see what are the changes that the original equation underwent to determine the new delta H.
Well, we can see one thing that happened is what was a product is now a reactant and what was a reactant are now products. So we reversed it. The next thing we should notice are the coefficients. Here it was a 2, but now it's a 4. This was a 4, but now it's an 8. This was a 3, but now it's a 6. So we didn't only reverse the reaction, but we multiplied by two.
Now remember when you reverse the reaction, that's going to reverse the sign of delta H, So now it's going to be positive. And whatever number I multiply my equation by, I multiply delta H by that same number. So you'd multiply this by two. When you do that, it gives you 5094 kilojoules. So for my new equation it would be a + 5094 kilojoules.
Reversing the reaction reverses the sign. Multiplying it by a value means I multiply my delta H value by that same number.