Equilibrium Constant 1) The Kc value for the following equilibrium at 500 degree

Question

Equilibrium Constant

1) The Kc value for the following equilibrium at 500 degrees Celcius is 49.

H2(g) + I2(g) = 2HI(g)

If 1.00 moles of H2 and 2.00 moles of I2 are introduces into a 3.00 L flask at 500 degrees celcius, how many moles of HI are present at equilibrium?

2) What would happen to the values you calculated for Kc in the experiment if the Beer's Law constant (k) value you used was low (e.g. you used 6000 M^-1 when the actual constant had a value of 10,000 M^-1) ? Explain your answer.

3) 1/2N2(g) + O2(g) = NO2(g) Kc = 1.04 x 10^-9

N2(g) + 2O2(g) = N2O4(g) Kc = 7.42 x 10^-18

Calculate Kc for each of the following.

(a) 2NO2(g) = N2(g) +2O2(g)

(b) N2(g) + 2O2(g) = N2O2(g)

(c) 2NO2(g) = N2O4(g)

Explanation / Answer

1) The Kc value for the following equilibrium at 500 degrees Celcius is 49.

H2(g) + I2(g) = 2HI(g)

If 1.00 moles of H2 and 2.00 moles of I2 are introduces into a 3.00 L flask at 500 degrees celcius, how many moles of HI are present at equilibrium?

Solution :- Lets first calculate the initial molarities of the H2 and I2

Molarity = moles / volume in liter

Molarity of H2 =1.00 mol / 3.00 L = 0.333 M

Molarity of I2 = 2.00 mol / 3.00 L = 0.667 M

Now lets make the ICE table for the reaction

       H2(g)     +   I2(g) ------- >    2HI(g)

I   0.333 M       0.667 M                 0

C   -x                      -x                      +2x

E   0.333 –x       0.667-x                 2x

Now lets write the Kc equation for the reaction

Kc =[HI]2/[H2][I2]

Lets put the values in the formula

49 = [2x]2/[0.333-x][0.667-x]

49 * [0.333-x][0.667-x] =[2x]2

49 * [0.333-x][0.667-x] =4x2

By solving this using the quadratic equation we get

x=0.33

now lets calculate equilibrium concentration of the HI

[HI] equilibrium = 2x =2*0.33 = 0.66 M

Therefore equilibrium concentration of the HI = 0.66 M

2) What would happen to the values you calculated for Kc in the experiment if the Beer's Law constant (k) value you used was low (e.g. you used 6000 M^-1 when the actual constant had a value of 10,000 M^-1) ? Explain your answer.

Solution :- If the value of the beers law constant used was smaller than the actual value of the beers law constant then calculated equilibrium constant Kc value would be smaller than actual. Because they are directly proportional.

3) 1/2N2(g) + O2(g) = NO2(g) Kc = 1.04 x 10^-9

N2(g) + 2O2(g) = N2O4(g) Kc = 7.42 x 10^-18

Calculate Kc for each of the following.

Solution :- first equation is reversed and multiplied by 2 therefore when we calculate the Kc for the new equation then we need to take inverse of the original Kc value and then take square of it

Kc = [1/Kcoriginal]2

Kc = [1/1.04*10^-9]2

Kc = 9.25*10^17

(b) N2(g) + 2O2(g) = N2O4(g)

Solution :- this is the same equation as given in the original second equation therefore Kc remains same

Kc= 7.42 x 10^-18

(c) 2NO2(g) = N2O4(g)

Solution :- to get the desired equation we need to reverse the equation 1 and multiply it by2 and then add it with equation 2

Lets write new equations

2NO2(g) ----- > N2(g) + 2O2(g)            kc =[1/Kc original]2

N2(g) + 2O2(g) = N2O4(g) Kc = 7.42 x 10^-18

2NO2(g) ------- > N2O4(g)       Kc = [1/Kc original]2*7.42 x 10^-18

Lets calculate the Kc

Put the values in the Kc equation

Kc = [1/Kc original]2*7.42 x 10^-18

Kc= [1/1.04*10^-9]2 * 7.42 x 10^-18

Kc = 6.86

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.