Calculate the standard enthalpy of reaction for the reaction 2Na + 2H2O——> 2NaOH+ H2. Standard enthalpies of formation are -285.8 kJ/mol for H2O and -470.11 kJ/mol for NaOH.
Enthalpy of reaction can be found by using the sum of enthalpies of products minus the sum of reactants. Remember that the Heat of formation of elements in their standard states is 0 kJ/mol.
[2(-470.110kJ/mol+(0)]-[2(-285.8kJ/mol)+2(0)]= ?
Right right, thank you so much. It's been a long day so my brain was in a bit of a mushy state. Thanks again!
eqn 1.....H2 + 1/2 O2 ==> H2O dH = -285.8 kJ/mol
eqn 2.....Na + 1/2 O2 + 1/2 H2 ==> NaOH dH = -470.11
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Multiply the reverse of equn 1 by 2 and add to twice eqn 2. Remember to change the sign of dH when a rxn is reversed. Add the two dH values.
Post your work if you get stuck. Check to make sure the equation at the end of the one you want.
Well, calculating the standard enthalpy of reaction is like playing a puzzle game. So, let's get started!
We have the equation: 2Na + 2H2O -> 2NaOH + H2
Now, we need to look at the enthalpy changes for each substance in the equation.
For 2Na, we don't have a standard enthalpy of formation, but we know that it's in its standard state. So, its enthalpy change is 0 kJ/mol.
For 2H2O, the standard enthalpy of formation is -285.8 kJ/mol. Since we have 2 moles of water, the enthalpy change for 2H2O is -571.6 kJ.
For 2NaOH, the standard enthalpy of formation is -470.11 kJ/mol. Again, since we have 2 moles of NaOH, the enthalpy change for 2NaOH is -940.22 kJ.
Lastly, for H2, we don't have a standard enthalpy of formation, but we can look up its bond energy, which is 436 kJ/mol.
Now, let's sum up the enthalpy changes:
Enthalpy change = (Enthalpy change of products) - (Enthalpy change of reactants)
= (2 x enthalpy change of NaOH) + (enthalpy change of H2) - (2 x enthalpy change of H2O)
Substituting in the values:
Enthalpy change = (2 x -470.11 kJ/mol) + (436 kJ/mol) - (2 x -285.8 kJ/mol)
= -940.22 kJ/mol + 436 kJ/mol - (-571.6 kJ/mol)
= -940.22 kJ/mol + 436 kJ/mol + 571.6 kJ/mol
= 67.38 kJ/mol
So, the standard enthalpy of reaction for this reaction is approximately 67.38 kJ/mol. Voilà!
To calculate the standard enthalpy of reaction (∆H°) for a reaction, you need to find the difference between the standard enthalpies of formation of the products and the reactants.
Given:
Standard enthalpy of formation (∆H°f) of H2O = -285.8 kJ/mol
Standard enthalpy of formation (∆H°f) of NaOH = -470.11 kJ/mol
The equation for the reaction is: 2Na + 2H2O -> 2NaOH + H2
Step 1: Determine the standard enthalpy of formation for each compound in the equation.
- The standard enthalpy of formation for Na (sodium) is zero because it is in its standard state.
- The standard enthalpy of formation for H2 (hydrogen gas) is also zero because it is in its standard state.
Step 2: Calculate the change in enthalpy for the reaction (∆H°) using the ∆H°f values.
∆H° = (2 * ∆H°f of NaOH) + (∆H°f of H2) - (2 * ∆H°f of Na) - (2 * ∆H°f of H2O)
∆H° = (2 * -470.11 kJ/mol) + (0 kJ/mol) - (2 * 0 kJ/mol) - (2 * -285.8 kJ/mol)
∆H° = -940.22 kJ/mol + 0 kJ/mol - 0 kJ/mol + 571.6 kJ/mol
∆H° = -368.62 kJ/mol
Therefore, the standard enthalpy of reaction for the given equation is -368.62 kJ/mol.