1. For the reaction 2 NO (g) 2H2 (9)N2(9) 2H20 (9) ined that, at equilibrium at

Question

1. For the reaction 2 NO (g) 2H2 (9)N2(9) 2H20 (9) ined that, at equilibrium at a particular temperature, the concentrations are as follows: 2(g)] = 5.3 × 102 M, and [H2O(g)] = 2.9 x 10" M. .3 [NO(g)] = 8.1 × 103 M, [H2(g)] = 4.1 x 10-5 M, [N Calculate the value of K for the reaction at this temperature. 2. At a particular temperature, a 3.0-L flask contains 2.4 moles of Cl2, 1.0 mol of NOCI, and 4.5 × 10-3 moles of NO. Calculate K at this temperature for the following reaction 2NOCI (g2 NO (9)+Cl2 (9)

Explanation / Answer

NO(g) + 2H2(g) ----------> N2(g) + 2H2O(g)

    Kc   = [N2][H2O]^2/[NO]^2[H2]^2
         =   5.3*10^-2 *(2.9*10^-3)^2/(8.1*10^-3)^2 *(4.1*10^-5)^2
     Kc   = 4*10^6

2. molarity of NO [NO] = no of moles/volume in L
                          = 4.5*10^-3/3 = 0.0015M
   molarity of Cl2 [Cl2] = no of moles/volume in L
                          = 2.4/3 = 0.8M
   molarity of NOCl      = no of moles/volume in L
                         = 1/3   = 0.33M
        2NOCl(g) ----------> 2NO(g) + Cl2(g)
         
           Kc = [NO]^2[Cl2]/[NOCl]^2
               = (0.0015)^2*(0.8)/(0.33)^2
               = 1.65*10^-5

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