Athletes use cold packs containing an ammonium nitrate solution to ice their inj
ID: 505635 • Letter: A
Question
Athletes use cold packs containing an ammonium nitrate solution to ice their injuries. 1.25 grams of ammonium nitrate is dissolved in enough water to form 25.0mL of a solution in a calorimeter. The temperature of the solution falls from 25.8 degree C to 21.9 degree C. What is the enthalpy of the reaction in kJ/mole? Assume the density the solution to be 1.0g/mL and its specific heat capacity to be 4.18 J/g degree C. (assume the calorimeter heat capacity to be negligible) 2. Calculate delta H for the reaction CH_4 (g) + NH_3 (g) rightarrow HCN (g) + 3 H_2 (g), given: N_2 (g) + 3 H_2 (g) rightarrow 2 NH_3 (g) delta H = -91.8 kJ C(s) + 2 H_2 (g) rightarrow CH_4 (g) delta H = -74.9 kJ H_2 (g) + 2 C(s) + N_2 (g) rightarrow 2 HCN (g) delta H= +270.3 kJ Calculate delta H for the reaction 2 Al (s) + 3 Cl_2 (g) rightarrow 2 AlCl_3 (s), given: 2 Al (s) + 6 HCl (aq) rightarrow 2 AlCl_3 (aq) + 3 H_2 (g) delta H = -1049. kJ HCl(g) rightarrow HCl (aq) delta H = -74.8 kJ H_2 (g) + Cl_2 (g) rightarrow 2HCl (g) delta H = -1845. kJ AlCl_3 (s) rightarrow AlCl_3 (aq) delta H = -323. kJExplanation / Answer
2. For the given data,
divide first equation by 2, and invert the reaction,
NH3(g) ---> 1/2N2(g) + 3/2H2(g) dH = 91.8/2 = 45.9 kJ
Invert the second equation,
CH4(g) ---> C(s) + 2H2(g) dH = 74.9 kJ
divide the third equation by 2,
1/2H2(g) + C(s) + 1/2N2(g) --> HCN(g) dH = 270.3/2 = 135.15 kJ
Add the three equations,
CH4(g) + NH3(g) ---> HCN(g) + 3H2(g) dH = 45.9 + 74.9 + 135.15 = 255.95 kJ
3. Multiply second equation by 6,
6HCl(g) --> 6HCl(aq) dH = 6 x -74.8 = -448.8 kJ
multiply third equation with 3,
3H2(g) + 3Cl2(g) ---> 6HCl(g) dH = 3 x -1845 = -5535 kJ
multiply third equation with 2 and invert the reaction,
2AlCl3(aq) ---> 2AlCl3(s) dH = 2 x 323 = 646 kJ
Add all to the first equation,
2Al(s) + 3Cl2(g) --> 2AlCl3(s) dH = (-1049 + (6 x -74.8) + (3 x -1845) + (2 x 323))
= -6386 kJ
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