In a type of water-jacketed bomb calorimeter (C_cal = 11347 J/degree C), the spa

Question

In a type of water-jacketed bomb calorimeter (C_cal = 11347 J/degree C), the space inside the calorimeter is filled with 500.0 g of water (C_s = 4.18 J/g degree C). Inside the water there is a reaction vessel containing 35 g of dicyanoacetylene C_4N_2 and an excess of O_2 gas. The dicyanoacetylene was combusted according to the following balanced reaction: C_4N_2(l) + 11/2 O_2(g) rightarrow 4 CO_2(g) + NO(g) + NO_2(g) As the combustion occurred, the calorimeter increased in temperature by 14 degree C and the water increased in temperature by 6.5 degree C. a. Using the data above, determine Delta E of the combustion of dicyanoacetylene in kJ/mole. b. If the reaction above was carried out in a variable volume (constant pressure) container like a piston-cylinder, would the system work be negative (-) or positive (+)? Briefly explain. c. The standard enthalpy of formation (Delta H degree f) of dicyanoacetylene is known to be highly endothermic. Give the balanced reaction demonstrating the formation of dicyanoacetylene from pure materials in their standard states. d. Estimate the standard enthalpy of formation of dicyanoacetylene (Delta H degree f) given the following data:

Explanation / Answer

a.

dE for combustion is defined as:

Isolated so:

Ecalorimeter + Ewater + Ereaction = 0

so

Ecalorimeter + Ewater = -Ereaction

Ecalorimeter = C*dT

Ewater = m*Cw*dT

where m = mass of water, Cp = specific heat of caloriemter, Cw = specific heat of water, dT = change in Temperature in calorimeter

so

Ecalorimeter = C*dT = 11347 J/C * 6.5 C = 68082 J

Ewater = 500 g *4.184 J/gC * 14 C = 29288 J

Total heat absorbed = 68082+29288 = 97370 J

then

Eabsorbed = -Ereaction

Ereact = -97370 J ; since it is negative, it must be exothermic, and makes sense since the surroundings increased T

but this is J, we need kJ/mol

so change to kJ

Ereaction = -97370J =- 97.370 kJ

and we used 35 g of specie

mol = mass/MW

MW of specie 76.06 g/mol

mol = 35/76.06 = 0.46 mol of specie

then

Ecombustion = -97.370/0.46 = 211.673 kJ/mol

Ecombustion of C4N2 = -211.673 kJ/mol

b.

If this was carried at constant pressure, there will be an increase in volume, since:

1 mol of C4N2 liquid + 11/2 mol of O2 gas are converted to 4 + 1 + 1 = 6 mol of gas

so mol of gas increases, Volume must increase

the system is doing WORK TO the surroundings, it is loosing energy so the Work will be positive according to

Q - W = dE

where W is inlet/outlet of work

c.

C4N2(l) must be formed

then we need Carbon andNitrogen in its elemental states

so

C(s) for carbon and N2(g) for Nitrogen

so

C(s) + N2(g) = C4N2(l)

we need 4 mol of C in order to balance this equation

4C(s) + N2(g) = C4N2(l)

note that all are in pure states

d.

Calcualte dHf if:

4C(s) + N2(g) = C4N2(l)

Hf = Bonds formed - bonds destroyed

bonds formed = C4N2:

C(triple)N: 2 times

886*2 = 1772 kJ/mol

C-C: 2 times

346*2 = 692 kJ/mol

C(triple)C: 1 time

698*1 = 698 kJ/mol

Total for Bonds formed = 1772+692+698 = 3162 kJ/mol

for Bonds destroyed:

C(s) = 0

N2(g)

N(triple)N = 942*1 = 942 kJ/mol

Bonds destroyed energy = 942 kJ/mol

dHf = 3162-942 = 2220 kJ/mol for C4N2

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.