Question

Hi! I am doing a chemistry lab and I am lost on where to start.

Below are the equations of interest in this laboratory exercise. As you know from reading the laboratory handout, your goal is to determine the enthalpy change for reaction 1

1. (1) Mg(s) + 1/2O2(g) → MgO(s) ∆H1=∆Hf(MgO) = ?

(2) Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) ∆H2 (measured)

(3) MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(I) ∆H3 (measured)

(4) 1/2O2(g) + H2(g) → H2O(l) ∆H4 = -286.0 kJ/mol

To arrive at equation (1) from the others …

Equation (2) needs to be (pick one) kept the same reversed reversed and x2. Equation (3) needs to be (pick one) kept the same reversed reversed and x2. Equation (4) needs to be (pick one) kept the same reversed reversed and x2.

Write out the results from each of your 3 answers above and then add them together to verify that the manipulations you describe sum to the desired reaction …

Answer 1

Sol :-

To find the enthalpy of formation of Magnesium Oxide

Chemical Reactions:

1) Mg(s) + 2HCl → MgCl2(aq) + H2(g)

2) MgO(s) + 2HCl → MgCl2(aq) + H2(g)

3) H2 + ½ O2 → H2O

In order to find the enthalpy of formation of the following reaction:

Mg(s) + ½ O2 → MgO(s)

We must find the enthalpy change of all the above reactions and then addthem in the following manner to determine the enthalpy of formation ofmagnesium oxide.ΔHf˚ of MgO = (ΔHf of equation 1) - (ΔHf of equation 2) + (ΔHf of equation 3)

Heat energy from reaction = Heat absorbed by calorimeter + heatabsorbed by reaction mixture.

-n ΔH˚ of reaction = Ccal ΔT + (m) (s) (ΔT)Where;n = number of moles of product found from the limiting reagent

ΔH˚ = heat evolved by the reaction

Ccal= calorimeter constanti.e. (mass of beaker + stirrer) x specific heat of borosilicate glass (0.71kJ kg-1 K-1)

ΔT = temperature change resulting from the reaction

m = mass of the solution

s = specific heat of the solution

Data collected for My added to HCl(aq) :- lemperan 126/27/28/33 37 37 36 36 35 565 535 535 535-51 Data collected for MgO added to HCl (9): Time/min /1/2/3/4/5/6/7/8/9/10/11/12/131 Temperature 25 25 25/27/28/2929529-44-29-829/29/29/291 S.Ganna CamScanner

Value obtained = 590KJ/mol. (rounded off)

Literature value= 601KJ/mol.

% Difference = 1.767%

Due to the errors, the value obtained is a little less than the literature value.

Hence, the enthalpy change for reaction 1 = $\Delta Hf$ =601KJ/mol