3.00 mol of PCl5 is placed in a 1L vessel and allowed to reach equilibrium at 380K. If Kc = 1.80 for

Q: What is the correct answer for CBSE11?

The correct answer is B. Let x be the amount of PCl5 dissociated. Initial [PCl5] = 3.00 M. At equilibrium: [PCl5] = (3.00 - x) M, [PCl3] = x M, [Cl2] = x M. Kc = [PCl3][Cl2] / [PCl5] = x^2 / (3.00 - x) = 1.80. This gives x^2 + 1.80x - 5.40 = 0. Using the quadratic formula: x = (-1.80 ± sqrt(1.80^2 - 4*1*(-5.40))) / 2 = (-1.80 ± sqrt(3.24 + 21.6)) / 2 = (-1.80 ± sqrt(24.84)) / 2 = (-1.80 ± 4.984) / 2. Taking the positive r

Q: 3.00 mol of PCl5 is placed in a 1L vessel and allowed to reach equilibrium at 380K. If Kc = 1.80 for PCl5(g) PCl3(g) + Cl2(g), what is the equilibrium concentration of PCl3?

The correct answer is B. Let x be the amount of PCl5 dissociated. Initial [PCl5] = 3.00 M. At equilibrium: [PCl5] = (3.00 - x) M, [PCl3] = x M, [Cl2] = x M. Kc = [PCl3][Cl2] / [PCl5] = x^2 / (3.00 - x) = 1.80. This gives x^2 + 1.80x - 5.40 = 0. Using the quadratic formula: x = (-1.80 ± sqrt(1.80^2 - 4*1*(-5.40))) / 2 = (-1.80 ± sqrt(3.24 + 21.6)) / 2 = (-1.80 ± sqrt(24.84)) / 2 = (-1.80 ± 4.984) / 2. Taking the positive r

Q: What are the options for this CBSE11 question?

A. 1.41 M · B. 1.59 M · C. 3.00 M · D. 0 M

Question

3.00 mol of PCl5 is placed in a 1L vessel and allowed to reach equilibrium at 380K. If Kc = 1.80 for PCl5(g) <=> PCl3(g) + Cl2(g), what is the equilibrium concentration of PCl3?

FirstInTest · Accepted Answer

Let x be the amount of PCl5 dissociated. Initial [PCl5] = 3.00 M. At equilibrium: [PCl5] = (3.00 - x) M, [PCl3] = x M, [Cl2] = x M. Kc = [PCl3][Cl2] / [PCl5] = x^2 / (3.00 - x) = 1.80. This gives x^2 + 1.80x - 5.40 = 0. Using the quadratic formula: x = (-1.80 ± sqrt(1.80^2 - 4*1*(-5.40))) / 2 = (-1.80 ± sqrt(3.24 + 21.6)) / 2 = (-1.80 ± sqrt(24.84)) / 2 = (-1.80 ± 4.984) / 2. Taking the positive root: x = (3.184) / 2 = 1.59 M. So, [PCl3] = 1.59 M. This is directly from Problem 6.9 in the document.

Answer

1.41 M

Answer

1.59 M

Answer

3.00 M

Answer

0 M

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