Calculate the change in heat:
2 Na(s) + 2 H2O(l) > 2 NaOH(aq) + H2(g)
So far this is what I have:
2 H20 --> 2 H2 + O2 2(285.83) kJ
2 Na + O2 + H2 --> 2 NaOH 2(-426.73) kJ
2 Na + 2 H2O --> 2 NaOH + H2 -281.8 kJ
If this correct??
yes, correct, as best as I can tell.
To calculate the change in heat for a chemical reaction, you need to consider the enthalpy values (ΔH) for each individual reaction and combine them appropriately.
In your example reaction, you've correctly identified the two separate reactions:
1) 2 H2O(l) → 2 H2(g) + O2(g) with a ΔH of 2(285.83) kJ
2) 2 Na(s) + O2(g) + H2(g) → 2 NaOH(aq) with a ΔH of 2(-426.73) kJ
To calculate the overall change in heat for the given reaction, you need to ensure that the reactants and products align. This can be achieved by reversing the second equation.
Reversed equation for the second reaction:
2 NaOH(aq) → 2 Na(s) + O2(g) + H2(g) with a ΔH of -2(-426.73) kJ
Now, add the two equations together:
2 H2O(l) → 2 H2(g) + O2(g) with a ΔH of 2(285.83) kJ
2 NaOH(aq) → 2 Na(s) + O2(g) + H2(g) with a ΔH of -2(-426.73) kJ
Overall balanced equation:
2 H2O(l) + 2 NaOH(aq) → 2 Na(s) + 2 H2(g) + O2(g)
Now, add the enthalpies:
ΔH = 2(285.83) kJ + (-2(-426.73) kJ)
ΔH = 571.66 kJ + 853.46 kJ
ΔH = 1425.12 kJ
Therefore, the correct change in heat for the given reaction is -1425.12 kJ.