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Elan Coil [88]
2 years ago
5

PCl5 decomposes to PCl3 and Cl2. If the initial concentration of PCl5 is 0. 200 m and the final concentration is 0. 120 m. What

is the equilibrium constant for the reaction?
Chemistry
1 answer:
lina2011 [118]2 years ago
3 0

PCl5 decomposes to PCl3 and Cl2.At an equilibrium constant is 0.0533.

The equilibrium constant, ok, expresses the connection among merchandise and reactants of a reaction at equilibrium with admire to a selected unit. The quantity values for are taken from experiments measuring equilibrium concentrations. The value of k shows the equilibrium ratio of merchandise to reactants. In an equilibrium mixture each reactants and products co-exist. big k > 1 products are desired k = 1 neither reactants nor merchandise are preferred.

The equilibrium will shift within the reverse path so that the effect of increased strain is nullified. for this reason, with growth in stress, the dissociation of PCl5 is suppressed and the degree of dissociation of PCl5 decreases.

PCl5  ===========> PCl3 +  CL2

0.200                                   0             0    (initial)

0.200-x                                x             x   (at equilibrium)

at equilibrium [PCL5] = 0.120

SO,

0.200 - x = 0.120

x = 0.080 M

Kc = [PCl3] [Cl2] / [PCl5]

    = x*x / (0.200 -x)

     = 0.080 * 0.080 / (0.120)

      = 0.0533

Hence equilibrium constant is  = 0.0533.

Learn more about concentration here:-brainly.com/question/17206790

#SPJ4

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What volume of oxygen (in L) is produced
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Answer:

12.36 L.

Explanation:

We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:

Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]

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Mole of Pb(NO₃)₂ = 147.1 / 331.22

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3 years ago
Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
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Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

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P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

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1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

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∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

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