Question

The combustion reaction described in part (b) occurred in a closed room containing 5.56 X 10' g of air

originally at 21.7°C. Assume that all of the heat produced by the reaction was absorbed by the air
(specific heat = 1.005 J/(g. °C)) in the room.
(c) Determine the final temperature of the air in the room after the combustion.
One method of producing ethanol is by the reaction of ethene and water, as represented below.
C2H4(8) + H2O(g) = C2H5OH(g)
AH° = 45 kJ/molna
(d) A chemist wants to run the reaction and maximize the amount of C2H5OH(g) produced. Identify two ways
the chemist could change the reaction conditions (other than adding or removing any chemical species)
to favor the formation of more product. Justify your answer.

Answer 1

The final temperature of the air in the room after the combustion can be determined using the formula: q = mcΔT. By substituting the values into the formula, we can calculate the final temperature of the air in the room after the combustion.

Step-by-step explanation:

The final temperature of the air in the room after the combustion can be determined using the formula:

q = mcΔT

Where q is the heat absorbed by the air, m is the mass of the air, c is the specific heat capacity of air, and ΔT is the change in temperature.

We know that the heat absorbed by the air is equal to the heat produced by the combustion reaction, which can be calculated using the heat of combustion of ethanol. The heat of combustion of 1 mole of ethanol can be calculated by subtracting the enthalpy of formation of water and carbon dioxide from the enthalpy of formation of ethanol:

ΔH = ΔHf(C2H5OH) - ΔHf(H2O) - ΔHf(CO2)

Once we have the heat absorbed by the air, we can rearrange the formula to solve for the final temperature:

ΔT = q / (m * c)

By substituting the values into the formula, we can calculate the final temperature of the air in the room after the combustion.

Answer 2

Step-by-step explanation:

Given that :

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g) H = −1270 kJ/mol

b) When a sample of C2H5OH was combusted, the volume of CO2(g) produced was 18.0 L when measured at 21.7°C and 1.03 atm. Determine the number of moles of CO2(g) that was produced.

SOLUTION:

Given that:

The volume of CO2(g) produced = 18.0 L

Temperature = 21.7° C = ( 21.7 + 273.15) K = 294.85 K

Pressure = 1.03 atm

Using the ideal gas equation:

PV = nRT

where R = constant = 0.0821 L .atm/K mol

1.03 × 18.0 = n × 0.0821 × 294.85

18.54 = n × 24.207185

n = 18.54 / 24.207185

n = 0.766 mole

Therefore, number of moles of CO2 (g) formed = 0.766 mole

(b_2) Determine the amount of heat, in kJ, that is released by the combustion reaction. The combustion reaction occurred in a closed room containing 5.56 × 104 g of air originally at 21.7°C. Assume that all of the heat produced by the reaction was absorbed by the air (specific heat = 1.005 J/(g °C)) in the room.

the number of moles of CO2 that is burned is : 0.766/2 mol

1 mole of C2H5OH = -1270 kJ/mol

0.766/2 mol of C2H5OH = -1270 × 0.766/2

= -486.34 kJ

Thus; the amount of heat released = -486.34 kJ

(c) Determine the final temperature of the air in the room after the combustion.

Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ = ΔT + T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

(d) A chemist wants to run the reaction and maximize the amount of C2H5OH(g) produced. Identify two ways the chemist could change the reaction conditions (other than adding or removing any chemical species)

to favor the formation of more product. Justify your answer.

C₂H₄ (g) + H₂O(g) -------> C₂H₅OH (g) ΔH = -45 kJ/mol

two ways the chemist could change the reaction conditions to favor the formation of more product either by :

1. Decreasing the temperature because the reaction is already exothermic ( i.e ΔH = -45 kJ/mol) ; so by decreasing the temperature, it will make the reaction to shift to the forward reaction which will in turn favor the formation of more products.
2. Increasing the pressure will initiate the movement of the reaction to proceed to the direction with lesser moles of gaseous product which will eventually favors the product side.