Abstract: solution. Next, the magnesium oxide and HCl

Abstract:

Calorimetry is used to find the enthalpy of a MgO solution.

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Going through the procedure properly, the enthalpy of the formation for MgO
will be found.

This will be done by first determining the heat of the first
and second reaction. The known value for formation of water will be plugged in
the equation to give the enthalpy of MgO. The goal of this experiment is to us
Hess’s Law to  calculate rxn  using the equation reaction= products,-  reactants  .

 

 

 

Introduction:

The main purpose of this
experiment is to find the heat of formation for magnesium. The reaction and
heat between Magnesium and hydrochloric acid is measured and recorded. To find
the heat given off by these reactions the equation used in coffee
cup calorimetry initially takes the initial and final temperature of the water
to find the amount of heat that is being absorbed or released during the
reaction. The formula used to find the heat is . As m is the mass of the solution, s is the specific heat
and is the final temperature- the initial temperature of the
solution. Next, the magnesium oxide and HCl is mixed and the heat  in the reaction is recorded again using the
formula for coffee calorimetry. This method used to measure the heat of the
reactions, coffee cup calorimetry is designed to measure the heat lost or
gained, as mentioned before. Although Hess’s Law has the equation reaction =
products- reactants, another rule it has is by adding all the enthalpy changes
for all the steps is the same as the overall enthalpy of a reaction. Coffee
Calorimetry also has a useful formula although Hess’s Law is used, another way
of finding the specific heat is the formula .

 

 

 

 

 

 

 

 

 

Data: 

Table 1: Initial and Final Temperatures of Mg

 

 

Initial Mass (g)

Minimum temperature (Celsius)

Maximum temperature (Celsius)

Trial 1

0.50

21.5

56.6
=+35.1

Trial 2

0.55

21.7

60.48
 =+38.7

 

Table 2: Initial and Final Temperatures of MgO

 

Initial Mass (g)

Minimum temperature (Celsius)

Maximum temperature (Celsius)

Trial 1

0.846

21.7

32.4

 
 

Trial 2

.927

21.6

33.3

 

Table 2- Trial 2 Initial and Final Temperatures of Mg and MgO

 

Results:  

 

rxn Trial 1
(kJ/mol)

rxn Trial 2
(kJ/mol)

rxn Average for 2 trials
(kJ/mol)

4(kJ/mol)

Mg

 

 

 

 

MgO

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Discussion: 

The experiment went smoothly, the procedure was successful
and the results were able to help accomplish the goal of finding the amount of
heat released in the reaction. There may have been a few sources of error but
the most effecting source that may have caused error is measurement. Getting
more than necessary of a solution is the most likely. Although the measurements
read to be correct, transferring solutions from one thing to another may also
affect the amount of solution. Not cleaning the measuring equipment properly may
lead to contamination; that also may have affected the outcome.

 

For an exothermic reaction the temperature rises due to heat
release to the surroundings of the system and H is negative. For reaction 1 and
2 the reactions are exothermic because the H number is negative, which
indicates the heat is being released from the system. In this lab quantity or q
is different from H°rxn, the definition of q is how much heat is lost or gained
in a system (chemwiki).The definition for H is a thermodynamic quantity equivalent to
the total heat content of a system. It is equal to the internal energy of the
system plus the product of pressure and volume.(google def of enthalpy)  H = E + PV. H and q are only similar when it
comes to constant pressure.

 

 

Further Analysis 

 

A.)  The
first step is to obtain the mass of the alloy, then dissolve the alloy in the
HCl solution. Then, one would need to take out the non reacted metal and weigh
it. This finds the mass of Mg by taking the whole mass of HCl and adding the
mass of the alloy.

B.)  The
percent would be calculated by multiplying the answer by 100. If the sample is
30% then it would be 5g multiplied .3, which equals 1.5grams of Mg, then it would
be multiplied by 24 mol of Mg to get .0625.