sodium metal reacts violently with water to produce aqueous sodium hydroxide and hydrogen gas. calculate the enthalpy change for this reaction.
is it:
Na + H2O -> ????
Hf is hafnium.
i don't know this stuff.
2Na + H2O ==> H2 + 2NaOH
Hf is Hf = heat formation and hafnium is not involved with anything your are doing.
Look up the delta Hf values (they should be listed in your text as delta Hof then
follow the equation I gave you. DHf stands for delta Hof.
I made a typo trying to type the sub subscripts and superscripts.
.....should be listed in your text as delta Hof then follow......
i balanced the equation:
2 Na + 2 H2O = H2 + 2 NaOH.
What delta Hf value am i suppose to look up?
Well, I'm not exactly a chemistry expert, but I can tell you that sodium and water definitely know how to make a splash!
As for the enthalpy change, I'm afraid I can't do the math for you. But hey, don't be too salty about it! Chemistry can be quite exothermic, and things might heat up when sodium takes a dip in water. So, let's just say it has a "fiery" reaction.
To calculate the enthalpy change for a reaction, we need to use the concept of heat of formation and Hess's Law.
First, we need to identify the balanced chemical equation for the reaction:
2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)
Next, we can look up the standard enthalpy of formation (ΔHf) values for each substance involved in the reaction. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm).
ΔHf for NaOH(aq) = -469 kJ/mol
ΔHf for H2(g) = 0 kJ/mol
ΔHf for Na(s) = 0 kJ/mol
ΔHf for H2O(l) = -286 kJ/mol
Now, we can apply Hess's Law, which states that the overall enthalpy change of a reaction can be calculated by the sum of the enthalpy changes of the individual steps.
The enthalpy change for the given reaction will be:
ΔH = Σ(ΔHf products) - Σ(ΔHf reactants)
Calculating for the enthalpy change:
ΔH = (2 mol × ΔHf NaOH) + (1 mol × ΔHf H2) - (2 mol × ΔHf Na) - (2 mol × ΔHf H2O)
ΔH = (2 mol × -469 kJ/mol) + (1 mol × 0 kJ/mol) - (2 mol × 0 kJ/mol) - (2 mol × -286 kJ/mol)
ΔH = -938 kJ - 572 kJ
ΔH = -1510 kJ
Therefore, the enthalpy change for the given reaction is -1510 kJ.
Write the equation and balance it.
Look up Hf for the products and reactants.
Then DHrxn = (n*DHfproducts) - (n*DHfreactants)