Julie is conducting an experiment where she placed 24.5 mL of water in a calorim
ID: 893821 • Letter: J
Question
Julie is conducting an experiment where she placed 24.5 mL of water in a calorimeter at 13.0 degrees C. Then, 2.0 g of A (molar mass = 48.0 g/mol), also at 13.0 degrees C, added to the water in the calorimeter and the temperature of the solution increases to 30.0 degree C. The following reaction is produced:A(s)rightarow A(aq) What is the enthalpy of the reaction ( Delta H degrees rxn), in kj/mol? Using the following informing: B(s) rightarrow B(aq) Delta H degrees rxn =+23.8 kj/mol 2B(aq) rightarrow A(aq) Delta Delta H degrees rxn =-1.25 kj /mol calculate Delta H degrees rxn for the following reaction (in lk/mol). A(s) rightarrow 2B(s) Delta H degrees rxn = -Explanation / Answer
V = 24.5
T = 13
m = 2g
MW = 48
Tf = 30
Q = m*Cp*(Tf-Ti)
Q = 24.5*4.18*(30-13) = 1740 J
Q = dHrxn = 1740 J
mols of rxn = 2/48 = 0.04166
Hrxn = 1740/0.04166 = 41766 J = 41.766 kJ
b)
B(S) --> B(aQ) H = 23.8 kJ
2B(aq) --> A(aq) H = 1.25 kJ
A(s) --> A(q) H = 41.766 kJ
Apply hess theorem
A(s) --> 2B(s)
Move the next equations
Invert B(S) --> B(aq) H = 23.8 kJ
B(aq) -->B(S) H = -23.8 kJ
2X by two
2B(aq) -->2B(S) H = -2*23.8 kJ = -47.6
Invert next one 2B(aq) --> A(aq) H = 1.25 kJ
A(aq) --> 2B(aq) H = -1.25 kJ
Leave alone this one A(s) --> A(q) H = 41.766 kJ
Total addition
2B(aq) -->2B(S) H = -2*23.8 kJ = -47.6
A(aq) --> 2B(aq) H = -1.25 kJ
A(s) --> A(q) H = 41.766 kJ
A(s) --> 2B(s) H = -47.6 + -1.25 +41.766 = -7.084 kJ/rxn
H = -7.084 kJ/rxn
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