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Question: Calc the $eq^m$ conc. of $PCl_3$ in 10L rigid vess of 1 mole of $PCl_5$ is ______ to attain the foll...

Calc the eqmeq^m conc. of PCl3PCl_3 in 10L rigid vess of 1 mole of PCl5PCl_5 is ______ to attain the following reactn^n.

PCl5(g)⇌PCl3+Cl3(g)PCl_5(g) \rightleftharpoons PCl_3 + Cl_3(g)

(i) if kc=10−5k_c = 10^{-5} at temp T1T_1 if kc=10+5k_c = 10^{+5} at temp T2T_2 if kc=10−2k_c = 10^{-2} at temp T3T_3

Answer

(i) 10^{-3} M, (ii) 0.1 M, (iii) 0.027 M

Explanation

Solution

The reaction is PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g).

Initial moles: PCl5=1PCl_5 = 1, PCl3=0PCl_3 = 0, Cl2=0Cl_2 = 0 in a 10L vessel. Let xx be the moles of PCl5PCl_5 that dissociate at equilibrium.

Equilibrium moles: PCl5=1−xPCl_5 = 1-x, PCl3=xPCl_3 = x, Cl2=xCl_2 = x.

Equilibrium concentrations (Volume = 10L): [PCl5]=1−x10[PCl_5] = \frac{1-x}{10} M [PCl3]=x10[PCl_3] = \frac{x}{10} M [Cl2]=x10[Cl_2] = \frac{x}{10} M

The equilibrium constant KcK_c is given by: Kc=[PCl3][Cl2][PCl5]=(x10)(x10)(1−x10)=x210(1−x)K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{(\frac{x}{10})(\frac{x}{10})}{(\frac{1 - x}{10})} = \frac{x^2}{10(1 - x)}

We need to solve for xx for each given value of KcK_c and then calculate [PCl3]=x10[PCl_3] = \frac{x}{10}. The equation is Kc=x210(1−x)K_c = \frac{x^2}{10(1 - x)}, which can be rearranged to x2+10Kcx−10Kc=0x^2 + 10 K_c x - 10 K_c = 0. The valid solution for xx must be in the range 0≤x≤10 \le x \le 1.

(i) If Kc=10−5K_c = 10^{-5}: The equation is x2+10(10−5)x−10(10−5)=0  ⟹  x2+10−4x−10−4=0x^2 + 10(10^{-5})x - 10(10^{-5}) = 0 \implies x^2 + 10^{-4}x - 10^{-4} = 0. Since KcK_c is very small, the reaction extent xx will be small (x≪1x \ll 1). We can approximate 1−x≈11-x \approx 1. Kc≈x210K_c \approx \frac{x^2}{10} 10−5≈x210  ⟹  x2≈10−4  ⟹  x≈10−4=10−210^{-5} \approx \frac{x^2}{10} \implies x^2 \approx 10^{-4} \implies x \approx \sqrt{10^{-4}} = 10^{-2}. Since x=10−2=0.01x = 10^{-2} = 0.01 is much smaller than 1, the approximation is valid. Equilibrium concentration of PCl3PCl_3 is [PCl3]=x10≈10−210=10−3[PCl_3] = \frac{x}{10} \approx \frac{10^{-2}}{10} = 10^{-3} M.

(ii) If Kc=10+5K_c = 10^{+5}: The equation is x2+10(105)x−10(105)=0  ⟹  x2+106x−106=0x^2 + 10(10^5)x - 10(10^5) = 0 \implies x^2 + 10^6 x - 10^6 = 0. Since KcK_c is very large, the reaction proceeds almost to completion, so xx is close to 1. Let x=1−yx = 1 - y, where yy is small (y≪1y \ll 1). Kc=(1−y)210yK_c = \frac{(1-y)^2}{10y}. Since yy is small, (1−y)2≈12=1(1-y)^2 \approx 1^2 = 1. 105≈110y  ⟹  106y≈1  ⟹  y≈10−610^5 \approx \frac{1}{10y} \implies 10^6 y \approx 1 \implies y \approx 10^{-6}. So, x=1−y≈1−10−6=0.999999x = 1 - y \approx 1 - 10^{-6} = 0.999999. Equilibrium concentration of PCl3PCl_3 is [PCl3]=x10≈1−10−610=0.1−10−7[PCl_3] = \frac{x}{10} \approx \frac{1 - 10^{-6}}{10} = 0.1 - 10^{-7} M. This is very close to 0.1 M.

(iii) If Kc=10−2K_c = 10^{-2}: The equation is x2+10(10−2)x−10(10−2)=0  ⟹  x2+0.1x−0.1=0x^2 + 10(10^{-2})x - 10(10^{-2}) = 0 \implies x^2 + 0.1x - 0.1 = 0. Using the quadratic formula: x=−0.1±(0.1)2−4(1)(−0.1)2=−0.1±0.01+0.42=−0.1±0.412x = \frac{-0.1 \pm \sqrt{(0.1)^2 - 4(1)(-0.1)}}{2} = \frac{-0.1 \pm \sqrt{0.01 + 0.4}}{2} = \frac{-0.1 \pm \sqrt{0.41}}{2}. Since x>0x>0, we take the positive root: x=−0.1+0.412x = \frac{-0.1 + \sqrt{0.41}}{2}. 0.41≈0.6403\sqrt{0.41} \approx 0.6403. x≈−0.1+0.64032=0.54032=0.27015x \approx \frac{-0.1 + 0.6403}{2} = \frac{0.5403}{2} = 0.27015. Equilibrium concentration of PCl3PCl_3 is [PCl3]=x10≈0.2701510=0.027015[PCl_3] = \frac{x}{10} \approx \frac{0.27015}{10} = 0.027015 M.

Summary of equilibrium concentrations of PCl3PCl_3: (i) Kc=10−5K_c = 10^{-5}: [PCl3]≈10−3[PCl_3] \approx 10^{-3} M (ii) Kc=10+5K_c = 10^{+5}: [PCl3]≈0.1[PCl_3] \approx 0.1 M (iii) Kc=10−2K_c = 10^{-2}: [PCl3]≈0.027[PCl_3] \approx 0.027 M