[smileshaikh]

Well I know to calculate heat flow you use, q=ms delta t.

for this experiment we mixed naoh with hcl into water.

the total mass (volume) of the mixture was 100g

the temperature change for naoh was 11.7 degrees C and for hcl it was 11.8 degrees C

now I'm looking to calculate heat flow

so so far I've got

q= (100.0g h20)(4.18 j/g degrees C) (?? ) do I put the hcl or the naoh, or the average of both temperatures combined... or am I supposed to calculate twice, once for the hcl and once for the naoh?

thanks.

[Unknown008]

Did you do the experiment? If so, could you tell me the details of how you measured the change in temperature for NaOH and HCl?

[smileshaikh]

Did you do the experiment? If so, could you tell me the details of how you measured the change in temperature for NaOH and HCl?

Sure, and we did three trials.

Here is all of the info for trial 1:

trial 1:

volume of 2 M NaOH used: 50.0 mL
initial temperature of NaOH: 24.4 degrees C
volume of 2 M HCl used: 50.0 mL
initial temperature of HCl: 24.3 degrees C
final temperature reached: 36.1 degrees C
total mass (volume) of mixture: 100.0 g
temperature change, delta t: NAOH: 11.7 degrees C, HCl 11.8 degrees C
Heat flow, joules:??
Moles of water produced:??

delta H (kj, mol water)??
mean value of delta H:??

the ?'s I haven't got to yet. But this is all the info for each of the three trials, of couse for the other two they are slightly different.

[smileshaikh]

Sure, and we did three trials.

Here is all of the info for trial 1:

trial 1:

volume of 2 M NaOH used: 50.0 mL
initial temperature of NaOH: 24.4 degrees C
volume of 2 M HCl used: 50.0 mL
initial temperature of HCl: 24.3 degrees C
final temperature reached: 36.1 degrees C
total mass (volume) of mixture: 100.0 g
temperature change, delta t: NAOH: 11.7 degrees C, HCl 11.8 degrees C
Heat flow, joules: ?????
Moles of water produced: ????

delta H (kj, mol water) ????
mean value of delta H: ?????

the ?'s i havent got to yet. but this is all the info for each of the three trials, of couse for the other two they are slightly different.

For the change in temperature of the NaOH and the HCL, I used the following method:

for the NAOH, I subtracted 36.1 degrees C- 24.4 degrees C= 11.7 degrees C NaOH

and for the HCl, I subracted 36.1 degrees C- 24.3 degrees C = 11.8 degrees C HCl

[smileshaikh]

As for how we measured it. We just poured 50.0 mL of each substance in a graduated cylinder.

[Unknown008]

Hmm, okay, the method is:

You take the initial temperature as the average of the two initial temperatures.

Hence, average initial temperature = (24.4 + 24.3)/2 = 24.35 C.

From there, you calculate your temperature change, and you plug that in your formula.

Post what you get for the other parts!

I hope it helped! :)

[smileshaikh]

Hmm, okay, the method is:

You take the initial temperature as the average of the two initial temperatures.

Hence, average initial temperature = (24.4 + 24.3)/2 = 24.35 C.

From there, you calculate your temperature change, and you plug that in your formula.

Post what you get for the other parts!

I hope it helped! :)

Thanks so much. I've calculated and got 4932.0 J for trial 1

I'm doing moles of water produced now. This has to do with the molecular mass? So I'd get the molecular mass of naoh+hcl?

Naoh+hcl but it was inside water too

[smileshaikh]

so what I'm doing to calculate heat moles of water produced is (50.0 mL NaOH)(1 L/ 1000 mL) (2 Mol/ 1 L) (1 Mol H2O/ 1 Mol NaOH)= 0.15 mol H2O

is this right? And what about for HCL because it wasn't even used?

[smileshaikh]

Hmm, okay, the method is:

You take the initial temperature as the average of the two initial temperatures.

Hence, average initial temperature = (24.4 + 24.3)/2 = 24.35 C.

From there, you calculate your temperature change, and you plug that in your formula.

Post what you get for the other parts!

I hope it helped! :)

so what I'm doing to calculate heat moles of water produced is (50.0 mL NaOH)(1 L/ 1000 mL) (2 Mol/ 1 L) (1 Mol H2O/ 1 Mol NaOH)= 0.15 mol H2O

is this right? And what about for HCL because it wasn't even used?

so would there be two answers for each trial? Or was I supposed to incorporate HCL in that formula I used and didn't do it?

[Unknown008]

To find that, you need to write the equation.



How many moles of NaOH did you have?
And how many moles of HCl did you have?

From there, use your equation to determine the number of moles of water produced.

Number of moles of NaOH =

Number of moles of HCl =

Hence, all the NaOH and HCl react to produce 0.1 mol of water.

[smileshaikh]

To find that, you need to write the equation.



How many moles of NaOH did you have?
And how many moles of HCl did you have?

From there, use your equation to determine the number of moles of water produced.

Number of moles of NaOH =

Number of moles of HCl =

Hence, all the NaOH and HCl react to produce 0.1 mol of water.

Thanks a lot unknown, you're a great help. I did just that.

[Unknown008]

Yes, that 0.1 mol of water produced the amount of energy you got above.

Find that for 1 mole to get the delta H of the reaction. :)

Repeat for the other trials and find the mean of all the delta H you obtained.

Glad it helped! :)

[smileshaikh]

Yes, that 0.1 mol of water produced the amount of energy you got above.

Find that for 1 mole to get the delta H of the reaction. :)

Repeat for the other trials and find the mean of all the delta H you obtained.

Glad it helped! :)

Oh it sure did, I got everything perfectly and sent it. You really are a great help to me and I appreciate it. I know it might get annoying every body asking for help all the time so that's why I like to thank you for your time and effort. You know, my math and chemistry is horrible and it takes me forever to comprehend so I try to do my best, as I want to be a med student one day.

[Unknown008]

If you have more questions you're now sure with, or have problems, feel free to post on the site. Be sure to post math problems in the right area though :)

Good luck! I hope you'll make your dream come true :)