Solving Enthalpy Questions: Calculating Heat Change in a Coffee-Cup Calorimeter

[eleventhxhour]

So I have two more enthalpy questions that I don't understand how to solve.

1) A student placed 50.0mL of 2.05mol/L NaOH in a coffee-cup calorimeter at 20.4°C. After quickly stirring the mixture, its temperature rose to 28.2°C. Determine the enthalpy change for the reaction:

2NaOH(aq) + H2SO4(aq) --> 2H2O(l) + Na2SO4(aq)

I tried converting NaOH and H2SO4 into grams, and then finding the q=mcdeltaT of it, but you don't have the c of the values so I couldn't do that.

 

[I like Serena]

So I have two more enthalpy questions that I don't understand how to solve.

1) A student placed 50.0mL of 2.05mol/L NaOH in a coffee-cup calorimeter at 20.4°C. After quickly stirring the mixture, its temperature rose to 28.2°C. Determine the enthalpy change for the reaction:

2NaOH(aq) + H2SO4(aq) --> 2H2O(l) + Na2SO4(aq)

I tried converting NaOH and H2SO4 into grams, and then finding the q=mcdeltaT of it, but you don't have the c of the values so I couldn't do that.

Hi eleventhxhour,

You need the enthalpy of each of those compounds.
You should have a table available in which you can find them.
For instance this table on wiki tells us that NaOH(aq) has a standard enthalpy change of formation of:
$$Δ_f H^0 = -470.1\text{ kJ/mol}$$

Since the unit is in kJ/mol, you will need the number of moles of each compound instead of the grams.

 

[eleventhxhour]

Hi eleventhxhour,

You need the enthalpy of each of those compounds.
You should have a table available in which you can find them.
For instance this table on wiki tells us that NaOH(aq) has a standard enthalpy change of formation of:
$$Δ_f H^o = -470.1\text{ kJ/mol}$$

Since the unit is in kJ/mol, you will need the number of moles of each compound instead of the grams.

So I converted each compound into moles and got 0.1025mol NaOH and 0.06molH2SO4. But I'm not sure what to do next. Would you use the equation q=mcdeltaT and find the total q for the reaction? And then use the equation deltaH=ndeltaHx? When I do this, I get the wrong answer...

 

[I like Serena]

So I converted each compound into moles and got 0.1025mol NaOH and 0.06molH2SO4.

Good!

But I'm not sure what to do next. Would you use the equation q=mcdeltaT and find the total q for the reaction? And then use the equation deltaH=ndeltaHx? When I do this, I get the wrong answer...

You are too early with your formula.
First you need to find the energy released by the reaction.

The standard enthalpy of formation is the amount of energy it takes to bind the atoms of 1 mole of the compound.
If you want a formula, it's:
$$q = n\cdot \Delta_f H^0$$

Note that this number is negative for $\ce{NaOH(aq)}$.
That should be expected, since energy would be released when in particular the O and H are bound together.

It means you can calculate the amount of energy it takes to form the compounds on the left hand side of the reaction.
And you can also calculate the amount of energy it takes to form the compounds on the right hand side.
The difference is the amount of energy that is released a part of the reaction.When you have that energy, you can apply your formula to calculate the temperature increase of the water.