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As we discovered in the lab. hydrochloric acid. HCI (aq), oxidizes magnesium met

ID: 483872 • Letter: A

Question

As we discovered in the lab. hydrochloric acid. HCI (aq), oxidizes magnesium metal to produce hydrogen gas and chloride ions. Magnesium ions are also produce Write a complete balanced chemical equation for the reaction described above. Determine the standard enthalpy change for this reaction. A 1.35 g ribbon of cleaned magnesium is dropped into a calorimeter containing 250.0 ml. of 0.50 M hydrochloric acid. If the initial temperature of the acid is 22.0 degree C, what is the final temperature of the solution? (Assume the density of the reaction solution is 1.0g/mL. the specific heat capacity is 4.184 J/g degree C, and that all the heat from the reaction is used to raise the temperature of the solution.) Is this reaction spontaneous at 25.0 degree C? At what temperature does this reaction become nonspontaneous?

Explanation / Answer

Q3.

a) the balanced reaction:

Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g)

note that MgCl2 is actually Mg2+ and 2Cl- in solution

b)

Enthalpy -->

HRxn = Hproducts - Hreactants

[1Hf(MgCl2 (aq)) + 1Hf(H2 (g))] - [1Hf(Mg (s)) + 2Hf(HCl (aq))]

[1(-801.15) + 1(0)] - [1(0) + 2(-167.15)] = -466.85 kJ

HRxn = -466.85 kJ     (exothermic)

c)

m = 1.35 g of Mg

mol = mass/MW = 1.35/24.305 = 0.0555 mol of Mg

m = 250 mL = 250 g of water

M = 0.5 M of HCl

mol HCL = MV = 0.250*0.5 = 0.125 mol of HCl

Ti = 22°C, find final...

limiting reactants --> 0.0555 mol o fMg requires --> 2*0.0555 = 0.111 mol , so Mg limits

Qreaction = mol*HRxn = 0.0555* 466.85 = 25.910175kJ

so

Qwater = m*C*(Tf-Ti)

25.910175*1000 = 250*4.184*(Tf-22)

Tf = 25910.175/250/4.184 + 22

Tf = 46.77 °C

d)

This must be spontanous, since dG < 0

for nonspotnaous:

dG = 0

so

dG = dH - T*dS

dH/dS = T

calculate entropy..

[1Sf(MgCl2 (aq)) + 1Sf(H2 (g))] - [1Sf(Mg (s)) + 2Sf(HCl (aq))]

[1(-25.11) + 1(130.59)] - [1(32.69) + 2(56.48)] = -40.17 J/K

-40.17 J/K     (decrease in entropy)

dH/dS = T

T = -466.85 / -40.17 = 11621.85 K

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