1) In a reaction, gaseous reactants form a liquid product. The heat absorbed by
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1) In a reaction, gaseous reactants form a liquid product. The heat absorbed by the surrounding is 26.0 kcal, and work done on the system is 15.0 Btu. Calculate Change in energy (Delta E) in kJ 2) A gas-producing reaction occurs in a container with piston- cylinder assembly u constant temperature, the volume increases from 10.5 L to 16.3 L. Calculate the work done (in J) 3) Nitroglycerine decomposes through an explosion that releases 5720 kJ of heat per mole. Is this reaction exothermic or endothermic? Draw an enthalpy diagram for this process C3H5NO3)3 (I)--d 3 CO2 (g) + 5/2 H2O (g) + ¼O2 (g) + 3/2 N2 (g) 4) In the following hydrogenation reaction, ethane (C2H4) and Hydrogen (H2) form ethane (C2H6). If 137 kJ is given off per mole of C2H4 reacting, How much heat is released when 15.0 Kg of C2H6 forms? C2H4 (g) +H2 (g) -à C2H6 (g) 5) Before baking a pie, you line the bottom of the oven with a 7.65 g piece of aluminum foil and then raise the temperature of oven from 18 C to 375 C. Find the heat absorbed (in J) by the A foil. (c of Al is 0.900 J/g.K) 6) A 12.18 g sample of a metal is heated to 65 C and then added to 25.00 g of water in a calorimeter. The water temperature changes from 25.55 C to 27.25 C. What is the unknown metal? 7) A 455 g piece of copper tubing is heated to 89.5 C and placed in an insulated vessel containing 159 g of water at 22.8 C. What is the final temperature of the system (c of water is 4.184 J/g.K, c of Cu is 0.387 J/g.K) 8) CO and NO are two pollutants that form in auto exhaust. A chemist must convert these pollutants to less harmful gases through the following process: CO (g) + NO (g) -d CO2 (g) + ½ N2 (g) Given the following information, calculate the unknown Delta H (Hint: use Hess's law) CO (g) + ½ O2 (g)-d CO2 (g) Delta H =-283 kJ N2 (g) 02 (g) --à 2NO (g) Delta H 180.6 kJ nder an external pressure o f 5.5 atm. AtExplanation / Answer
1btu=0.252Kcal
Work done=15Btu= 15*0.252 Kcal=3.78 Kcal
From 1st law of thermodynamics,
deltaU= Q+W
anything added to the system is +ve and leaving is –ve.
Accordingly, Q is –ve since heat is removed from system
And W is +ve. Work is added to system
deltaU= 3.78-26=-22.22 Kcal
1Kcal= 4.18 Kj
-22.22 Kcal= -22.22*4.18 KJ=-98.88 KJ
2. work done = PdV, work is done in the process of expansion is -ve
Work done = -5.5*(16.3-10.5) L.atm=-31.9 L.atm, 1L.atm=101.3 J, hence -31.9L.atm= -31.9*101.3 Joules=-3231.47 joules
3. since heat is releaased, the reaction is exothermic.
4. The reaction is C2H4(g)+H2(g)------->C2H6
1 mole of C2H4 gives 1 mole of C2H6
molar mass of C2H4= 28 and that of C2H6= 30
28 gm of C2H4 gives 30 gm of C2H6
30 gm of C2H6 137 KJ
15 kg =15*1000 gm of C2H6 gives 15*137*1000Kj=2055000 KJ
5. heat added to aluminium= mass of aluminium* specific heat* change in temperature= 7.65*0.9*(375-18) joules=2458 joules
6. heat removed from metal = heat added to water
mass of metal* specific heat of metal* change in the temperature= mass of water* specific heat of water* change in temperature
specific heat of water= 4.18 J/gm.deg.c
12.18*specific heat of metal*(65-27.25)= 25*4.18*(27.25-25.55)
specific heat of metal = 0.39 J/gm.deg.c. From specific heat data of metal, the metal is Zinc.
7. let the final temperature be t
hence heat removed from copper= heat added to water
mass of copper* specific heat of copper* change in temperature = mass of water* specific heat of water* change in temperature.
455*0.387*(89.5-t)= 159*4.184*(t-22.8)
176*89.5-176t= 665.3t- 665.3*22.8
r*(665.3+176)= 176*89.5+665.3*22.8 = 36.75 deg.c
8.
The 1st reaction is CO(g)+0.5O2(g)--àCO2(g), deltaH=-283KJ (1)
The 2nd reaction is N2(g)+O2(g)------>2NO(g), deltaH= 180.6 Kj
Reversing the reaction 2NO(g)->N2(g)+ O2(g), deltaH= -180.6 KJ
Dividing by 2, NO(g)--->0.5N2(g)+0.5O2(g), deltaH=-90.3 Kj (2)
Eq.1+Eq.2 gives CO(g)+NO(g)->CO2(g)+0.5N2(g), deltaH= -283-90.3=-373.3 KJ the enthalpy change of desired reaction.
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