Section: WORKSHOP 8 THERMOCHEMISTRY Shonw calculation setups and answers for eac
ID: 573994 • Letter: S
Question
Section: WORKSHOP 8 THERMOCHEMISTRY Shonw calculation setups and answers for each question. Please note that your instructor may opt to assign specific questions from those listed below LCalculate the change in internal energy (in ) for a balloon that is heated by adding 215 cal of Consider the following bala If the enthalpy change is 16.5 kJ. how many grams of hydrogen gas are produced? 2 ndreaction: CHOHg) COg) + 2H(s), where AH-907 kl A 50.00 g sample of an unknown substance absorbed 2.578 kJ of energy as it changed from a temperature of 250 C to 89.7 "C, What is the specific heat of this unknown substance (in lig CP g k An alloy of mass 250 g was heated to 88.6 and then placed in a calorimeter that co 61.2 g of water at 19.6 4. . The temperature of the water rose to 21.3 Determine the peck heat of the alloy (in J/g C).Explanation / Answer
1. Expansion of balloon
change in internal energy dE = heat given + work done
with,
heat = 215 cal x 4184 = 899560 J
work done = -422 J
So,
dE = 899560 - 422 = 899138 J
3. Specific heat of unknown substance = Cp
Cp = heat absorbed/mass x change in temerpature
feeding values from above,
Cp (unknown metal) = 2.578 x 10^3 J/50 g x (89.7 - 25) = 0.797 J/g.oC
4. Heat lost by hot metal = heat gained by water
heat = mCpdT
Cp = specific heat of metal
So,
25 g x Cp x (88.6 - 21.3) = 61.2 x 4.184 x (21.3 - 19.2)
Cp (metal) = 0.320 J/g.oC
6. dHo(rxn) calculation
Multiply first equation with 2,
invert the last equation
add all together,
2Mg(s) + O2(g) ---> 2MgO(s) dHo = 2 x -601.7 = -1203.4 kJ
Mg(s) + S(s) ---> MgS(s) dHo = -598 kJ
SO2(g) ----> S(s) + O2(g) dHo = 296.8 kJ
------------------------------------------------------------------------------------------------------------
3Mg(s) + SO2(g) --> MgS(s) + 2MgO(s) dHo(rxn) = -1504.6 kJ
7. First equation multiply by 2,
second equation invert and multiply by 2,
third equation multiply by 3
Add them together
2N2(g) + 2O2(g) --> 4NO(g) dHo = 2 x 43.20 kcal = 86.4 kcal
4NH3(g) ---> 2N2(g) + 6H2(g) dHo = 2 x 22.10 kcal = 44.2 kcal
6H2(g) + 3O2(g) ---> 6H2O(g) dHo = 3 x -115.60 kcal = -346.80 kcal
----------------------------------------------------------------------------------------------------------
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g) dHo(rxn) = -216.2 kcal