For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 2.89...

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For the following reaction, Keq is 0.00183 at 390. K: PCl5(g)PCl3(g) + Cl2(g) If 2.89 g of PCl5 is placed in a 2.86 L bulb at390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013bar
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3.8 Ratings (337 Votes)

PCl5(g)------------> PCl3(g) + Cl2(g)

no of moles of PCl5 = W/G.M.Wt

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.89/208 = 0.0139moles

PV = nRT

PÂ Â = nRT/V

Â Â Â Â = 0.0139*0.0821*390/2.86Â Â = 0.156atm

Â Â Â Â Â Â Â Â Â Â Â Â PCl5(g) ---------->PCl3(g) + Cl2(g)

IÂ Â Â Â Â Â Â Â Â Â Â 0.156Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â +x

EÂ Â Â Â Â Â Â 0.156-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

Â Â Â Â Â Â Â Â Â Â KpÂ Â = P_PCl3 P_Cl2/P_PCl5

Â Â Â Â Â Â Â Â Â 0.00183Â Â = x*x/0.156-x

Â Â Â Â Â Â Â Â Â 0.00183*(0.156-x) = x²

Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â = 0.016

Â Â Â Â Â Â Â Â P_{Xl2Â Â Â Â} = xÂ Â Â = 0.016atm = 0.016*1.013Â Â = 0.016208bar

Â Â Â Â Â

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For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 2.89...

Free

80.2K

Verified Solution

Question

Chemistry

For the following reaction, Keq is 0.00183 at 390. K: PCl5(g)PCl3(g) + Cl2(g) If 2.89 g of PCl5 is placed in a 2.86 L bulb at390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013bar
answer in bar

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For the following reaction, Keq is 0.00183 at 390. K: PCl5(g) PCl3(g) + Cl2(g) If 2.89...

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80.2K

Verified Solution

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Chemistry

For the following reaction, Keq is 0.00183 at 390. K: PCl5(g)PCl3(g) + Cl2(g) If 2.89 g of PCl5 is placed in a 2.86 L bulb at390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013baranswer in bar

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For the following reaction, Keq is 0.00183 at 390. K: PCl5(g)PCl3(g) + Cl2(g) If 2.89 g of PCl5 is placed in a 2.86 L bulb at390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013bar
answer in bar