Given the labeled reactions (A-D) from the lab description and the following exp

Question

Given the labeled reactions (A-D) from the lab description and the following experimental data, complete the following calculations of ?H for the reactions of Mg and MgO. Assume the reactions are carried out in a calorimeter containing 100 mL H2O.

A) H2(g) + 1/2O2(g) --> H2O(l) ?Hf= -68.3 kcal/mol

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

C) MgO(s) + 2HCl(aq) --> MgCl2(aq) + H2O(l)

D) Mg(s) + 1/2Os(g) --> MgO(s)

For reaction B: Initial temp: 21.0 C, Final temp: 31.8 C, Wt. of Mg ribbon: 0.2481 g ?H(B)= ? kcal/mol

For reaction C: Initial temp: 21.0 C, Final temp 24.1 C, Wt. of MgO ribbon: 0.4204 ?H(C)= ? kcal/mol

Using the data above, calculate the enthalpy of formation of MgO= ? kcal/mol

Explanation / Answer

B) heat dH = -q = mCpdT

with,

m = 100 ml x 1.00 g/ml = 100 g

Cp = 4.184 J/g.oC

dT = 31.8 - 21.0 = 10.8 oC

we get,

dH = -100 x 4.184 x 10.8 = -4.51872 kJ

dH = -4.51872 kJ/0.2481 g = -18.2133 kJ/g

dH(B) = -18.2133 kJ/g x 24.305 g/mol = -442.67 kJ/mol

C) heat dH = -q = mCpdT

with,

m = 100 ml x 1.00 g/ml = 100 g

Cp = 4.184 J/g.oC

dT = 24.1 - 21.0 = 3.1 oC

we get,

dH = -100 x 4.184 x 3.1 = -1.29704 kJ

dH = -1.29704 kJ/0.4204 g = -3.085 kJ/g

dH(C) = -3.085 kJ/g x 24.305 g/mol = -74.99 kJ/mol

D) enthalpy of formation of MgO,

invert C) and add A), B) and C),

H2(g) + 1/2O2(g) + Mg(s) + 2HCl(aq) + MgCl2(aq) + H2O(l) ---> H2O(l) + MgCl2(aq) + H2(g) + MgO(s) + 2HCl

cancel common terms,

Mg(s) + 1/2O2(g) ----> MgO(s)

dH for MgO = -68.3 - 442.67 + 74.99 = -435.98 kJ/mol

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