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3 years ago

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Calculate the number of moles of Cl2 produced at equilibrium when 3.98 mol of PCl5 is heated at 283.9 deg celsius in a vessel ha

ving a capacity of 10.0L. At 283.9, K = 0.060 for this dissociation.

1 answer:

ddd [48]3 years ago

7 0

Answer:

1.274 moles

Explanation:

The equation for the reaction can be represented as follows:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

K = 0.060

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Concentration of $PCl_5(g)$ = $\frac{numbers of moles}{volume}$

Concentration of $PCl_5(g)$ = $\frac{3.98}{10.0}$

Concentration of $PCl_5(g)$ = 0.398 moles

If we construct an ICE table for the above equation; we have:

$PCl_5(g)$ ⇄ $PCl_3(g)$ + $Cl_2(g)$

Initial 0.398 0 0

Change - x + x + x

Equilibrium (0.398 - x) x x

K = $\frac{[PCl_3][Cl_2]}{[PCl_5]}$

K = $\frac{[x][x]}{[0.398-x]}$

K = $\frac{x^2}{0.398-x}$

0.060 = $\frac{x^2}{0.398-x}$

0.06(0.398-x) = x²

0.02388 - 0.060x = x²

x² + 0.060x - 0.02388 = 0 (quadratic equation)

a = 1; b= 0.06; c= -0.02388

Using quadratic formula;

= $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

= $\frac{-0.06+/-\sqrt{(0.06)^2-4(1)(-0.02388)} }{(2*1)}$

= $\frac{-0.060+/-\sqrt{0.0036+0.09552} }{2}$

= $\frac{-0.06+/-\sqrt{0.09912} }{2}$

= $\frac{-0.06+/-0.3148}{2}$

= $\frac{-0.060+0.3148}{2}$ or $\frac{-0.060-0.3148}{2}$

= $\frac{0.2548}{2}$ or $\frac{-0.3748}{2}$

= 0.1274 or -0.1874

We go by the positive value which says:

[x] = 0.1274 M

number of moles = 0.1274 × 10.0

= 1.274 moles

∴ the number of moles of Cl₂ produced at equilibrium = 1.274 moles

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