Enhtalpy change – Hess’ law Essay Sample

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Introduction of TOPIC

Enhtalpy change – Hess’ law

Vanier college

Part A.

Objective:

The objective of this lab was to determine the standard heat formation (∆H°F) of MgO, using a calorimeter and determining the enthalpy of two reactions. Applying Hess’ law we were able to determine the standard heat formation of MgO.

Introduction

Energy exchanged in a chemical reaction can either be in the form of heat or light. If light is involved a glow is seen, if heat is involved the temperature of the system will change(lab manual page 35). The amount of heat exchanged under constant pressure is called the enthalpy change, this can either be endothermic or exothermic. Endothermic if heat is absorbed by the reaction (positive sign), and exothermic if heat is released by the reaction (negative sign). If a reaction is reversed, the sign of ΔH is also reversed. The heat change that results from the formation of one mole of a compound from its elements in the regular states is known as its standard heat of formation (∆°f). The specific heat of the solution is the amount of heat required to raise the temperature of 1 gram of the solution by 1°C. To find the enthalpy change of a reaction using a calorimeter we added the Mg and MgO solids to two HCl solutions in two separate Styrofoam cups, and using a thermometer and taking note of the temperature change every 30 seconds for 7.5 minutes we then plotted this on a graph.

Extrapolating the points from the final temperature after the 7.5 minutes, through where the heat of the reaction begins cooling, we were able to determine the maximum heat that would have been achieved if their was no heat loss in the reaction. Using this and the initial temperature we found ∆T for both of the reactions. We then calculated our masses of solutions by weighing precisely the amount of Mg and MgO used, as well as the masses of the HCl solutions. The specific heat of solution was given which was 4.184 J/g°C. Using the formula q=(∆T)(Specific heat of solution)(Mass of solution) we were able to determine the amount of heat released by the reactions. Knowing our limiting reagents this let us calculate the enthalpy changes of each reaction in Kj/mol. The enthalpy change for reaction 3 was given. Applying Hess’ law using all three equations we were able to determine the heat of formation (∆°f) of MgO in Kj/mol.

Experimental

Refer to lab manual pages 38-39, also refer to attached pre-lab.

Data/Results
Reaction of Mg + 2H Mg +H2

Initial temp: 19°C
Extrapolated temp: 60.8°C
Temperature difference (∆T): 41.8°C
Mass of Styrofoam cup: 3.533g
Mass of Styrofoam cup and solution: 64.194g
Mass of HCl solution: 60.661g
Quantity of heat released: 10.71680 Kj
Mass of Mg used: 0.616g
Amount of Mg used: MM: Mg = 24.31g/mol (.616g Mg)(1mol/24.31g Mg) =.0253 mols of Mg used.

This chart is of reaction temperature, over seven and a half minutes, of the equation Mg(s) + 2H(Aq)  Mg(Aq) + H2. We added Mg to our 2H+ and noted down the temperature change at every 30 seconds. We then extrapolated the values back from our final temperature through the point in where the cooling started. This made it possible for us to determine ∆T for the reaction.

∆T= Our initial temperature was 19°C and the extrapolated temperature was 60.8°C. Text. –Tint. = 41.8°C(∆T)

Mass of solution = We added the mass of the HCl solution with the mass of Mg used:

60.661g+0.616g =61.277g of solution Specific heat capacity was given as 4.184 J/g°C because

the solution is mostly water.

Quantity of heat released = ∆T × Mass of solution × Specific heat of solution:
q= (61.277g)(4.184J/g°C)(41.8°C) = 10,716.80J
10,716.80 ÷ 1000 = 10.71680 Kj = q solution
qrxn= −qsoln
qrxn= −10.71680 Kj
∆Hrxn = qrxn/mol of solid used
∆Hrxn= −10.71680Kj/.0253mols =−424 Kj/mol

Reaction of MgO + 2H  Mg + H2O

Initial temp: 19°C
Extrapolated temp: 31.45°C
Temperature difference: 12.45°C
Mass of Styrofoam cup: 3.546g
Mass of Styrofoam cup and solution: 63.990g
Mass of HCl: 60.444g
Quantity of heat released: 3.20119Kj
Mass of MgO used: 1.010g
Amount of MgO used: MM MgO: 24.305 + 16.00= 40.305 g/mol  (1.010g MgO)(1 mol/40.305g)= 0.025 mols of MgO used

This chart has the same legend as the previous one but it is for the reaction MgO(s) + 2H(Aq)  Mg(Aq) + H20(L). Again, extrapolating the values back made it possible to find maximum temperature reached and determine ∆T.

By using the same methods of calculations seen above we obtain the heat released for this reaction.

Quantity of heat released = ∆T × Mass of solution × Specific heat of solution:
q= (61.45g)(4.184J/g°C)(12.45°C) = 3,201.19J
3,201.19J÷ 1000 = 3.20119 Kj = q solution
qrxn= −qsoln
qrxn= −3.20119Kj
∆Hrxn = qrxn/mol of solid used
∆Hrxn= −3.20119Kj/0.025mols =−128 Kj/mol

Hess’ Law
Using Hess’ law we can use all three reactions to determine the heat of formation of magnesium oxide. Equation 1 releases 424 Kj of energy , 2 releases 128 Kj, and 3 releases 285 Kj . Because they are releasing this energy during the reaction they are exothermic and therefore are negative vaues. We use Hess’ law and combine these three reactions calculating them algebraically, for equation 2 we need to reverse the reaction for it to fit in our equation to use Hess’ law. To perform the reaction Mg(s) + ½02  MgO we need to use Hess’ law because if not it would give off an uncontrollable flame which would make it impossible to use a calorimeter. Some errors that may have occurred in our data would be; from the transferring of the Mg or MgO to the Styrofoam cup, we may have spilled some, from the heat loss to the surroundings seeing as they were only Styrofoam cups that weren’t covered, and also because of our human reflex that may have caused small error in temperature readings.

1.) Mg(s) + 2H(aq)  Mg(aq) +H2 ∆H= −424 Kj
2.) MgO(s) + 2H(aq)  Mg(aq) + H20 ∆H= −128 Kj
3.) H2(g) + ½O2(g)  H2O(L) ∆H= −285.5 Kj

Mg(s) + ½O2(g)  MgO(s) ∆H=???

1) Mg(s) + 2H(aq)  Mg(aq) +H2 ∆H= −424 Kj
Reversed:2) Mg(aq) +H2O(L)  MgO + 2H ∆H= 128 Kj
3) H2(g) + ½O2(g)  H2O(L) ∆H= −285.5 Kj

Mg(s) + ½O2(g)  MgO(s) ∆H= −581.5 Kj

Conclusion
After finding the ∆H of equations 1 and 2 we were able to apply Hess’ law to determine the standard heat of formation of MgO. The ∆H of reaction 1 was −424 Kj, for 2 ∆H= −128 Kj and for 3 (given value) ∆H= −285.5 Kj. After applying Hess’ law we found the the ∆H°F for the reaction Mg(s) + ½02  MgO was −582 Kj/mol. This tells us that in the formation of MgO there is 582 Kj of energy released.

Part B.

∆H.lattice > ∆H.hydration = endothermic process

The process of dissolving NH4Cl is endothermic. In order for the crystal solid to dissociate into ions, energy must be absorbed, in this case the lattice energy was overcome and the solid separated into ions and no solid was left. Knowing that the solid completely dissolved in the solution we are able to determine that the lattice energy is greater then the hydration energy meaning that the reaction is endothermic. After the dissociation of the NH4Cl the temperature of the water dropped from 16°C to 9°C this is because the ions are cooling down the water in an exothermic process, known as hydration energy. An example of a exothermic process is the burning of a candle, the candle is giving off heat to its surroundings. A good example of an endothermic process is photosynthesis. Plants absorb energy from the sun allowing them to produce food and grow.

Part C.

Energy exchanged in a chemical reaction can either be in the form of heat or light. If light is involved a glow is seen, if heat is involved the temperature of the system will change (lab manual page 35). During this experiment we observed that when the clear liquid luminol is mixed with potassium ferricyanide a white blue glow of light is created from the mixture. The energy transformation in this experiment creates this glow, it is created from the decay of a high energy intermediate to a lower energy product, the excess energy is seen visually as photons and produce the blue glow. When matter is changed there is always a transfer of energy. Whether it be in the form of heat or pressure. For matter to change states the particles either have to get “excited” or “calmed”. The more you heat particles the more excited they get. This will produce a gas if heated high enough. The more the particles are cooled the more they slow down and eventually if cooled enough arrive at a solid state.

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