3.00 , mol of PCl_5 kept in 1 , L closed vess | vedclass

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Question

(A) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$At $t=0$: $[PCl_5] = 3 \, M$,$[PCl_3] = 0$,$[Cl_2] = 0$At equil

Vedclass · Accepted Answer

$1.59 \, M, 1.59 \, M$

Vedclass · Answer

(A) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$At $t=0$: $[PCl_5] = 3 \, M$,$[PCl_3] = 0$,$[Cl_2] = 0$At equilibrium: $[PCl_5] = (3-x) \, M$,$[PCl_3] = x \, M$,$[Cl_2] = x \, M$$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = 1.80$$1.80 = \frac{x^2}{3-x}$$5.4 - 1.8x = x^2$$x^2 + 1.8x - 5.4 = 0$Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-1.8 + \sqrt{(1.8)^2 - 4(1)(-5.4)}}{2(1)}$$x = \frac{-1.8 + \sqrt{3.24 + 21.6}}{2} = \frac{-1.8 + \sqrt{24.84}}{2}$$x = \frac{-1.8 + 4.984}{2} = \frac{3.184}{2} \approx 1.59 \, M$Therefore,$[PCl_3] = 1.59 \, M$ and $[Cl_2] = 1.59 \, M$.

$3.00 \, mol$ of $PCl_5$ kept in $1 \, L$ closed vessel was allowed to attain equilibrium at $380 \, K$. If $K_C = 1.80$,calculate the equilibrium concentrations of $PCl_3$ and $Cl_2$ respectively.

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