Chemistry lab report about Heat of reaction

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Lab report Heat of Reaction

Introduction The purpose of the lab was to through experiments calculate the heat of reaction and heat of formation, and see if we got a realistic result according to facts.

We should see if Reaction 1 and equalled

We created reactions and then calculated their enthalpies.

Cheap custom essay on Chemistry lab report about Heat of reaction

Work plan Substances Material

Material NaOH (s) Beakers, 00ml

NaCl (aq) 0.50 M Thermometer

HCl (aq) 0.5 M Balance

Deionised water

Work plan Since the lab was pretty time consuming we decided to

Method divide the lab into lessons, one practical, where we should do all the actual lab-work, and one theoretical, were we should do all the calculations and evaluate our result.

The laboratory part of the experiment takes place in steps, or reactions.

Reaction 1

We began by weighing a dry beaker. Then we filled it with 00ml-deionised water, and recorded the temperature of the fluid. After that was done we hydrated the water by adding .07 grams of NaOH (s). We mixed it thoroughly with the water, and as it reacted the temperature rose. We recorded the highest temperature. (See Calculations).

Reaction

We began in the same way as in reaction 1, by weighing the beaker, but instead of using deionised water we used 100 ml of 0.5 M HCl (aq), and recorded it¡¯s temperature. After that we added 0. grams of NaOH (s). The temperature rose and we recorded the highest temperature.

Reaction

We weighed the 00 ml beaker and added 50 ml of 0.50 M HCl (aq) to it, and recorded the temperature. After that we added 50 ml of 0.50 M NaOH (aq). The temperature rose, and we recorded the highest temperature.

Calculations Reaction 1 NaOH (s)¨¤Na+ (aq) + OH- (aq)

m (beaker) 18.81 g.

m (NaOH) .07 g. = 0.0506666 moles

V 00 ml of water.

c (beaker) 0. x 4.1 J/g

c (water) 4.1 J/g

T1 0.¡ã

T .7¡ã

¦¤T= T1 - T = 1.8¡ã

To find the enthalpy we have to use the equation

V x c (water) x ¦¤T, which equals the enthalpy in Joules.

I put in my values (see above) in the equation

V x c (water) x ¦¤T = 00 x 4.1 x 1.8 = 1508.4 J

This is only the heat taken up by the water. Of course the beaker will take up some heat too. To find this we have to use this equation

c (beaker) x m (beaker) x ¦¤T, this too, equals the enthalpy in joules.

I put in my values in the equation

(0. x 4.1) x 18.81 x 1.8 = 08.7004 J

To get the total heat of reaction I have to plus these values together like this

1508.4 + 08.7004 = 1717.1 J

Since this heat is released, the ¦¤H is negative. ¦¤H is in this case -1717.1 J.

Reaction

NaOH (s) + H+(aq) + Cl- (aq)¨¤ Na+(aq) + Cl- (aq) + HO

m (beaker) 17.70g.

m (NaOH) 0. g. = 0.0481065 moles.

V 100 ml of 0.5 M HCl (aq)

c (beaker) 0. x 4.1 J/g

c (HCl (aq)) 4.1 J/g

T1 0.¡ã

T 5.0¡ã

¦¤T= T1 - T = 4.1¡ã

To find the enthalpy for this reaction, I just do the same as I did in reaction 1.

Total equation

(V x c (HCl (aq)) x ¦¤T) + (c (beaker) x m (beaker) x ¦¤T)

I put in my values

(100 x 4.1 x 4.1) + (0.x4.1 x 17.70 x 4.1) = (1717.) +

(47.116516) = 11.0165 J

Just as in reaction 1, heat is released. ¦¤H is therefore negative. ¦¤H = - 11.0165 J

Reaction

Na+(aq) + OH- (aq) + H+ (aq) + Cl- (aq) ¨¤

Na+(aq)+Cl- (aq)+ HO

m (beaker) 141.04 g.

V (NaOH) 50 ml, 0.5 M

V (HCl) 50 ml, 0.5 M

V total 100 ml

c (beaker) 0.x4.1 J/g

c (NaOH+HCl) 4.1 J/g

T1 0.8

T .

¦¤T= T1 - T = .1¡ã

I do the same as in reaction 1 and .

Total equation

(V x c (NaOH+HCl) x ¦¤T) + (c (beaker) x m (beaker) x ¦¤T)

I put in my values

(100 x 4.1 x .1) + (0. x 4.1 x 141.04 x .1) = (18.) + (67.0751) = 1665.751 J

Just like reaction 1 and , does reaction release heat, and its ¦¤H is therefore negative. ¦¤H = - 1665.751 J

Calculating the heat of formation

The heat of formation is the heat of reaction divided by the number of moles we used, or the enthalpy change when one mole of a compound is formed from its element.

In reaction 1 we used 0.0506666 moles of NaOH

The heat of reaction in this experiment was -1717.1 J

To convert this into heat of formation we need to divide it by the number of moles of NaOH used, like this

-1717.1/0.0506666 = - 80.58 J = - .806 kJ

In reaction we used 0.0481065 moles of NaOH. Since we used the sodium hydroxide to neutralize the hydrochloric acid, it¡¯s not certain that this is the number of moles used in the reaction. We need to determine which substance is the limiting one. To do this I need to calculate the number of HCl moles in the hydrochloric acid.

n = c x v I put in my values

n = 0.5 x 0.1 = 0.05 moles

There is about 0.05 moles of each substance, but we have a little more of the hydrochloric acid, which means that NaOH is the limiting substance and the substance we should use for the calculation.

To calculate the heat of formation, I¡¯ll do the same as I did for the first reaction.

- 11.0165 / 0.0481065 = - 887.4554 J =

- 88.74554 kJ

In reaction the is the same number of moles of both of the substances.

n = c x v I put in my values for reaction

n = 0.50 x 0.05 = 0.05 moles

To get the heat of formation I divide the heat of reaction with the number of moles.

1665.751 / 0.05 = 666.1815 J = 66.61815 kJ

To get the equation which we were to use to check the correctness we had to reverse reaction number one and add it to number two so

¨C 1 =

To see if my answers are correct, I check them in this way

To get the total heat of formation you have to take reaction (- 87.6 kJ) minus reaction 1 (- . kJ) equals reaction , like this

¦¤H - ¦¤H1 = ¦¤H

- 88.74554 ¨C (- .8058) = - 54.818611 kJ

The answer should be equal to ¨C66.616 kJ. This means that we are about 1. kJ away from real answer.

Discussion I think our result is acceptable, but pretty far from accurate. I think that when perform a lab like this, there are many different sources of errors, which are hard to eliminate. Heat will always be lost to the surroundings for example. Something that I think we could maybe have done better job at recording the starting temperature. We just left the thermometer in for about one minute. Maybe we should have left it there longer and we might have gotten an even better result. But at least we used a digital one, instead of a normal one, that must have given us a better result. Other than that, I can¡¯t really think of anything we could have done wrong, or something that we could have done in a better way.

Could there be any errors in the calculations? The specific heat content of water is 4.1 J/g. We where supposed to use the same value for the hydrochloric acid, but is that really 100 % correct? It there a better value you could use to get a better result?

I don¡¯t think this is a really important lab, I don¡¯t think this is something that you do very often since heat of formation is so well documented in books. But the lab really helped me to understand the different between heat of reaction and heat of formation, and how to calculate these.

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