Given that:
H2 (g) + F2 (g) ---> 2HF (g) H(rxn)= -546.6 kJ
2H2 (g) + O2 (g) ---> 2H2O (l) H (rxn)= -571.6 kJ
calculate the value for H(run) for...
2F2(g) + 2H2O (l) ---> 4HF (g) + O2 (g)
_____kJ?
-521.6kJ
What's H(run) mean?
Sorry i meant H(rxn) (of the reaction)
To calculate the value of ΔH° for the given equation:
2F2(g) + 2H2O(l) ---> 4HF(g) + O2(g)
you can use Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway and depends only on the initial and final states.
Step 1: First, we need to determine the intermediate steps required to get from the given equations to the desired equation.
Given equation 1:
H2(g) + F2(g) ---> 2HF(g) ΔH° = -546.6 kJ
Given equation 2:
2H2(g) + O2(g) ---> 2H2O(l) ΔH° = -571.6 kJ
We can see that the desired equation can be obtained by adding equation 1 and equation 2.
But before adding, we need to make sure the coefficients of equation 1 and equation 2 match the desired equation.
Step 2: Balancing the intermediate equations:
To balance equation 1, we need to multiply it by 2.
2 (H2(g) + F2(g) ---> 2HF(g)) ΔH° = -2 * 546.6 kJ
2H2(g) + 2F2(g) ---> 4HF(g) ΔH1° = -1093.2 kJ
Now, we can add equation 1 and equation 2 to get the desired equation:
ΔH2° = ΔH1° + ΔH°
= (-1093.2 kJ) + (-571.6 kJ)
= -1664.8 kJ
Therefore, the value of ΔH° for the reaction 2F2(g) + 2H2O(l) ---> 4HF(g) + O2(g) is -1664.8 kJ.