Question

In: Chemistry

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

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 at 390. K, what is the equilibrium pressure of Cl2? 1 atm = 1.013 bar

answer in bar

Expert Solution

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

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

= 2.89/208 = 0.0139moles

PV = nRT

P = nRT/V

= 0.0139*0.0821*390/2.86 = 0.156atm

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

I 0.156 0 0

C -x +x +x

E 0.156-x +x +x

Kp = P_PCl3 P_Cl2/P_PCl5

0.00183 = x*x/0.156-x

0.00183*(0.156-x) = x²

x = 0.016

P_Xl2 = x = 0.016atm = 0.016*1.013 = 0.016208bar

queen_honey_blossom answered 1 year ago

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