Given the thermochemical equation:
2 Al(s) + 3/2 O2(g) > Al2O3(s) ΔH = -95.7 kJ
find the heat of reaction, ΔH, for the following reaction:
2 Al2O3(s)> 4 Al(s) + 3 O2(g)
Select one:
a. -95.7 kJ
b. +95.7 kJ
c. -47.8 kJ
d. +47.8 kJ
e. +191.4 kJ
You have the reaction as shown. You want the reverse reaction and double that. So take dH that you have, change the sign (because of the reversal) and double it.
would it be a positive or a negative though?
it would be e...
yes.
You reversed the sign because the reaction is reversed. So that makes it positive. Then double that for e.
thank you !!!
To find the heat of reaction, ΔH, for the given reaction, we can use the stoichiometric coefficients and the ΔH value of the thermochemical equation. Here's how we can do it step by step:
1. Start with the given thermochemical equation:
2 Al(s) + 3/2 O2(g) > Al2O3(s) ΔH = -95.7 kJ
2. Notice that the given reaction is the reverse of the thermochemical equation. Therefore, the sign of the ΔH value will be inverted.
3. Since the given reaction is the reverse, we need to reverse the stoichiometric coefficients as well. This means that instead of:
2 Al(s) + 3/2 O2(g) > Al2O3(s)
we have:
Al2O3(s) > 2 Al(s) + 3/2 O2(g)
4. Now, we can determine the ΔH value for the reverse reaction by changing its sign. Therefore, the ΔH value for the given reaction is +95.7 kJ (the inverse of -95.7 kJ).
Hence, the correct answer is option b. +95.7 kJ.