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The equilibrium constant, K, for the following reaction is 2.3 x 10-4 at 300 °C: N2...

The equilibrium constant, K, for the following reaction is 2.3 x 10-4 at 300 °C: N2 (g) + C2H2 (g) ⇌ 2 HCN (g). Calculate the equilibrium concentration of HCN when 0.555 moles of N2 and 0.555 moles of C2H2 are introduced into a 0.500 L vessel at 300 °C.

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Expert Solution

he equilibrium constant, K, for the following reaction is 2.3 x 10-4 at 300 °C: N2 (g) + C2H2 (g) ⇌ 2 HCN (g). Calculate the equilibrium concentration of HCN when 0.555 moles of N2 and 0.555 moles of C2H2 are introduced into a 0.500 L vessel at 300 °C.

Concentration = moles/volume

                                              N2                         C2H2                        HCN

initial concentration           0.555/0.5= 1.11              0.555/0.5= 1.11         0

change                             -x                                         -x                         2x

equilibrium                     1.11-x                                    1.11-x                  2x

Ka= equilibrium constant for the reaction , N2(g) + C2H2(g) <----> 2HCN(g)

Ka= [HCN]2/ [C2H2][N2] =2.3*10-4

4x2/ (1.11-x)2= 2.3*10-4, taking square root   2x/(1.11-x)= 0.0152

2x= 0.0152*(1.11-x)

2x+0.0152x= 0.0152*1.11

2.0152x= 0.0152*1.11, x =0.0084

hence at equilibrium [N2] =1.11-0.0084=1.092= [C2H2] , [ HCN]= 2*0.0084= 0.0168M

queen_honey_blossom answered 1 year ago
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