Question

1 Hydrogen and chlorine react to produce hydrogen chloride (HCl) as shown below. At equilibrium, the flask contains 0.239 g of HCl, 0.254 g of Cl₂ and 0.00013 g of H₂. Calculate the value of K_c.

The reaction is: H₂ + Cl₂ <--> 2HCl

Hint: Work in mol/L.

		1.72 x 103
		5.4 x 10-2
		1.84 x 102
		3.55  x 10-5
		2.99 x 104 2.If 40 g of HF are initially placed in an empty 1 L flask, calculate the amount of HF remaining if at equilibrium 1.0 g of H2 had been produced. The reaction is: 2HF  --> H2  +  F2 Hint:  Work in moles and then switch back to grams at the end. 20 g of HF 25 g of HF 39 g of HF 40 g of HF 1.0 g of HF

Answer 1

no of moles of HCl = W/G.M.Wt
                   = 0.239/36.5   = 0.00655moles
      [HCl]         = 0.00655mole/1 L   = 0.00655mole/L
no of moles of Cl2 = W/G.M.Wt
                    = 0.254/71 = 0.00357 moles
       [Cl2]        = 0.00357mole/L   = 0.00357mole/L
no of moles of H2 = W/G.M.Wt
                    = 0.00013/2   = 0.000065 moles
        [H2]        = 0.000065mole/1L   = 0.000065mole/L

       H2 + Cl2 <--> 2HCl
       
         Kc   = [HCl]^2/[H2][Cl2]
              = (0.00655)^2/0.00357*0.000065
              = 184
              = 1.84*10^2 >>>>answer
2.
       no of moles of HF = W/G.M.Wt
                          = 40/20   = 2 moles
      no of moles of H2 = W/G.M.Wt
                         = 1/2   = 0.5 moles
      2HF --> H2 + F2
I      2       0      0
C    -2*0.5    0.5    0.5
E      1       0.5    0.5  

   no of moles of Hf at equilibrium = 1moles
mass of HF at equilibrium = 1*20 = 20g

20 g of HF >>>>answer