Sometimes it is necessary to use the quadratic equation to find the value of the

Question

Sometimes it is necessary to use the quadratic equation to find the value of the variable x. You need to combine like terms in the form ax^2 + bx + c = 0 x can take two values x = -b + square root b^2 - 4 ac/2a or x = -b - square root b^2 - 4 ac/2a Only one of the two values of x will make sense - the other will either be negative or higher than the given initial concentration or not make sense in some other way. PCl_5 decomposes into PCl_3 and Cl_3. What are the []_eq if 0.10 M of each substance is present at the beginning of the reaction? K_c = 3.81 Times 10^2. Write the chemical equation, the expression of equilibrium constant (k_c) in terms of reactants and products, setup your ICE table and show all calculations.

Explanation / Answer

PCl5 <-> PCl3 + Cl2

K = [PCl3][Cl2] / [PCl5]

K = 3.81*10^2

initially

[PCl3]= 0.1

[Cl2]= 0.1

[PCl5]= 0.1

in equilibrium

[PCl3]= 0.1 + x

[Cl2]= 0.1 + x

[PCl5]= 0.1 - x

Substitutein K

K = [PCl3][Cl2] / [PCl5]

3.81*10^2 = (0.1 + x)(0.1 + x)/(0.1 - x)

solve for x

(0.1 + x)(0.1 + x) = (3.81*10^2)(0.1) - (3.81*10^2)*x

0.1^2 + 2*0.1*x + x^2 = 3.81*10  - (3.81*10^2)*x

(0.01-3.81*10 ) + (0.2 + 3.81*10^2)*x + x^2 = 0

x^2 + 381.2*x - 38.09= 0

a = 1; b = 381.2 ; C = -38.09

solve for x

x = ( -b + sqrt(b^-4ac) ) / 2a = (-381.2 + sqrt((381.2^2)-4*(1)(-38.09)))/2 = 0.0998951232

x = ( -b - sqrt(b^-4ac) ) / 2a =  (-381.2 - sqrt((381.2^2)-4*(1)(-38.09)))/2 = -381.29

choose the positive one

[PCl3]= 0.1 + x = 0.1+0.0998951232 = 0.199895

[Cl2]= 0.1 + x = 0.1+0.0998951232 = 0.199895

[PCl5]= 0.1 - x = 0.1-0.0998951232 = 0.000104

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