Q1 - The equilibrium constant, K c , for the following reaction is 7.00×10 -5 at

Question

Q1 - The equilibrium constant, Kc, for the following reaction is 7.00×10-5 at 673 K.

NH4I(s) NH3(g) + HI(g)  

Calculate the equilibrium concentration of HI when 0.295 moles of NH4I(s) are introduced into a 1.00 L vessel at 673K.  

[HI] =   M

Q2 - The equilibrium constant, Kc, for the following reaction is 9.52×10-2at 350 K.

CH4(g) + CCl4(g) 2 CH2Cl2(g)

Calculate the equilibrium concentrations of reactants and product when 0.286 moles of CH4and 0.286 moles of CCl4are introduced into a 1.00 L vessel at 350 K.

Q3- The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K.

2CH2Cl2(g) CH4(g) + CCl4(g)  

Calculate the equilibrium concentrations of reactant and products when 0.340 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K.  

Explanation / Answer

               NH4I (s) -------> NH3(g) + HI(g)

Initially      0.295/1 mol/L      0         0

change          -x                +x        +x

At equili     (0.295-x)           x         x

Kc = [HI][NH3]/[NH4I]

7 * 10^-5 = x*x/(0.295-x)

x = [HI] = 4.51*10^-3

                CH4(g) + CCl4(g) --------> 2CH2Cl2(g)

initially     0.286/1 mol/L 0.286            0

change         -x          -x               +2x

At equili   (0.286-x) (0.286-x)             2x

Kc = [CH2Cl2]^2/[CH4][CCl4]

9.52 * 10^-2 = (2x)^2/((0.286-x)*(0.286-x))

x = 0.0382

[CH4] = 0.286 - 0.0382 = 0.2478

[CCl4] = 0.286 - 0.0382 = 0.2478

[CH2Cl2] = 2(0.0382) = 0.0764

                  2CH2Cl2(g) -------> CH4(g) + CCl4(g)

Initially       0.34                    0            0

Change       -2x                     +x      +x

At equilib (0.34-2x)                   x     x

Kc = [CH4][CCl4]/[CH2Cl2]^2

10.5 = x*x/(0.34-2x)^2

[CH4] = [CCl4] = 0.201

[CH2Cl2] = 0

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