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  • Reaction between N_2 and O_2– takes place as follows: 2N_2 (g) + O_2 (g) 2N_2O (g) If a mixture of 0.482 mol N_2 and 0.933 mol of O_2 is placed in a 10 L

7.8      Reaction between N2 and O2-takes place as follows:

                2N_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2N_{2}O_{(g)} 

If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc = 2.0 ×10-37, determine the composition of equilibrium mixture. 

Answers (1)

best_answer

It is given that,
K_c = 2.0\times 10^{-37}
Let the concentration of N_2O at equilibrium be x. So,

                                2N_{2}_{(g)}+O_{2}_{(g)}\rightleftharpoons 2N_{2}O_{(g)}
initial conc                  0.482        0.933                     0          (in moles)
at equilibrium             0.482-x   0.933 -  x              x           (in moles)

The equilibrium constant is very small. So, we can assume   0.482-x = 0.482 and 0.933 -  x = 0.933

We know that,
 

K_c = \frac{[N_2O]^2}{[N_2]^2[O_2]}
2\times 10^{-37}=\frac{(x/10)^2}{(0.0482)^2(0.0933)}       {dividing the moles by 10 to get concentration of ions)

\\\frac{x^2}{100}= 2\times 10^{-37}\times (0.0482)^2\times (0.0933)\\ x^2 = 43.35\times 10^{-40}\\ x = \sqrt{43.35\times 10^{-40}}\\ x=6.6\times 10^{-20}

So, the concentration of [N_2O]= \frac{x}{10}=6.6\times 10^{-20}

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manish

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