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Oksana_A [137]
2 years ago
5

A sample of flammable liquid is placed into an enclosed cylinder which is then fitted with a movable piston. Initially the cylin

der contains a volume of 8.20 L. The sample is ignited producing gas and releasing 427.8 J of energy. To what volume will the cylinder expand to if it must expand against a pressure of 826.1 mmHg. Assume all the energy released is converted to work used to push the piston?
760 mmHg = 1 atm
101.3 J = 1 L atm
Chemistry
1 answer:
polet [3.4K]2 years ago
5 0

Answer:

12.09 L

Explanation:

Step 1: Convert 826.1 mmHg to atm

We will use the conversion factor 760 mmHg = 1 atm.

826.1 mmHg × 1 atm/760 mmHg = 1.087 atm

Step 2: Convert 427.8 J to L.atm

We will use the conversion factor 101.3 J = 1 L.atm.

427.8 J × 1 L.atm/101.3 J = 4.223 L.atm

Step 3: Calculate the change in the volume

Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:

w = P × ΔV

ΔV = w/P

ΔV = 4.223 L.atm/1.087 atm = 3.885 L

Step 4: Calculate the final volume

V₂ = V₁ + ΔV

V₂ = 8.20 L + 3.885 L = 12.09 L

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Answer:

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Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

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2. Derive the rate law

Assume k₋₁ ≫ k₂.  

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In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

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