Question

#6 The following reaction shows combustion reaction of propane gas at 25 Celsius.

**C _{3}H_{8(gas)}+5O_{2}_{(gas}_{)<===>}
3CO_{2(gas)+}4H_{2}0_{(liq)};
deltaH=-2219.97 kJ/mol**

Initially 1L-chamber contained 2.03M of propane
(C_{3}H_{8}), 3.50 M of oxygen (O_{2}) and
0.030M of carbon dioxide (CO_{2}). After equilibrium is
reached, the chamber contained 1.53 M of propane
(C_{3}H_{8}).

a)Using the information given,calculate equlibrium constant
(K_{c}) of thid reaction.

b) After equilibrium is reached, if the chamber size is increased to 5L, predict the direction of equilibrium shift. Explian your answer.

c)After equilibrium is reached, if we increase temperature to 1000 Celsius, predict the direction of equilibrium shift. Explain your answer.

Answer #1

a) C3H8(gas) + 5O2(gas) <------> 3CO2(g) + 4H2O(l)

Kc = [CO2]^3/[C3H8][O2]^5

at equillibrium

[ C3H8 ] = 2.03 -x

[ O2] = 3.50 - 5x

[ CO2] = 0.030 + 3x

Kc = (0.030 + 3x)^3/(2.03 - x) (3.50 - 5x)^5

at equillibrium

[C3H8 ] = 1.53M

2.03 - x = 1.53M

x = 0.5

Therefore,

Kc = (0.030+3x)^3/(2.03 - x) (3.50-5x)^5

= (0.030 + 3(0.5))^3/(2.03 - 0.5 ) ( 3.50 -5(0.5))^5

= 3.58/ 1.53

= 2.34

b) Increasing volume shift the equillibrium to the side which higher mole of gaseous product , reactant side having more moles (6 moles ( 1C3H8 + 5O2) ) than the product side (3mol (3CO2)).So, the equillibrium shift to the reactant side.

c) This reaction is exothermic reaction ( heat is released in this reaction).Exothermic reaction is not favourable for product side. So , the equillibrium shift reactant side.

Consider the combustion reaction of propane gas: C3H8(g) +
5O2(g) → 4H2O(g) + 3CO2(g) Predict the signs (+ positive, -
negative, 0 zero, or CBD cannot be determined ), if possible, for
the delta Ssystem. Use the symbol to fill the blank.

Consider the balanced equation for the combustion of propane,
C3H8 C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) If propane reacts with
oxygen as above a. what is the limiting reagent in a mixture
containing 5.00 g of C3H8 and 10.0 g of O2? b. what mass of the
excess reagent remains after the reaction ? c. what mass of CO2 is
formed when 1.00 g of C3H8 reacts completely?

15. Consider the following combustion reaction of propane. 3C3H8
+ 5O2 → 3CO2 + 4H2O
a) What mass of O2 , in g, would be needed to react with 0.421 kg
of C3H8?
b) What mass of CO2 , in g, would be produced from the combustion
of 0.421 kg of C3H8 with excess oxygen?

LP gas burns according to the following exothermic reaction:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)ΔH∘rxn=−2044kJ What mass of LP gas is
necessary to heat 1.5 L of water from room temperature (25.0 ∘C) to
boiling (100.0 ∘C)? Assume that during heating, 14% of the heat
emitted by the LP gas combustion goes to heat the water. The rest
is lost as heat to the surroundings

Liquefied petroleum (LP) gas burns according to the following
exothermic reaction:
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)ΔH∘rxn=−2044kJ
.
Part A
What mass of LP gas is necessary to heat 1.8 L of water from
room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that,
during heating, 14% of the heat emitted by the LP gas combustion
goes to heat the water. The rest is lost as heat to the
surroundings.
Express your answer using two significant figures.

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