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a mixture of 0.2 mol of NO, 0.1 mol of H2 and 0.2 mol of H2O...

a mixture of 0.2 mol of NO, 0.1 mol of H2 and 0.2 mol of H2O is placed in 2.0 L at 300 K. At equal [NO] = 0.062. Calculate K

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Expert Solution

Answer – We are given, moles NO = 0.20 moles , moles of H2 = 0.10 moles

Moles of H2O = 0.2 moles , volume = 2.0 L

Reaction

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

[NO] = 0.20 moles / 2.0 L =0.10 M

[H2] = 0.1 moles / 2.0 L = 0.05 M

[H2O] = 0.20 moles / 2.0 L = 0.10 M

At equilibrium, [NO] = 0.062 M

We need to put Ice chart

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

I   0.1            0.05            0               0.10

C -2x             -2x            +x             +2x

E 0.062      0.05-2x         +x              0.10+2x

We know,

At equilibrium, [NO] = 0.062 M = 0.10-2x

So, 2x = 0.10-0.062

             = 0.038

So, x = 0.019

So at equilibrium, [H2] = 0.05-2x

                                      = 0.05-2*0.019

                                       = 0.012 M

[N2] = x = 0.019 M

[H2O] = 0.1+2x

           = 0.1+0.038

            = 0.138 M

So, K = [N2 (g)] [H2O(g)]2 / [NO(g)]2 [H2(g)]2

        = (0.019) (0.138)2 / (0.062M)2(0.012)2

        = 654

queen_honey_blossom answered 2 years ago
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