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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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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

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Here, R is radius and is related to a as follows:

R=\frac{a}{2}

Putting the value in expression for area,

A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

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Since, 1 nm=10^{-7}cm

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Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of Cr^{3+} and O^{2-} is 62 pm and 140 pm respectively.

Converting them into cm:

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Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

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Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

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