Answer to Question #86515 in General Chemistry for Richa

Question

at 2000 degree celsius the equilibrium constant for reaction 2NO equals N2 + O2 is kc= 2400. if the initial concentrations of NO is 0.175 M. What are equilibrium concentrations of NO, N2, and O2.

Answer 1

$$2NO\leftrightarrow N_2 +O_2$$

$$K_c = \frac{[N_2][O_2]}{[NO]^2}$$

The equlibrium constant K_c is large, so we should srart with as much products as possible. If all of 0.0175 M of NO reacts to form products the concentrations would be:

$[NO] = 0 M$

$[N_2] = \frac{0.175 M}{2} = 0.0875 M$

$[N_2] = \frac{0.175 M}{2} = 0.0875 M$

Using these "shifted" values as initial concentartions with x as NO concentration at equilibrium gives the folliwing ICE table:

Since we are starting close to equilibrium, x should be small so that:

$0.0875 -0.5x \approx 0.0875$

$$K_c = \frac{(0.0875)(0.0875)}{x^2} = 2400$$

$x = 0.00179 M$

Select the smallest concentartion for the 5% rule:

$$\frac{0.00179}{0.0875}\times100\% = 2.05\%$$

This value is less than 5%, so the assumptions are valid.

The concentrations at equilibrium are:

$[NO] = x = 0.00179 M$

$[N_2] = 0.0875 - 0.5x = 0.0875 - 0.5\times 0.00179 = 0.0866 M$

$[H_2] = 0.0875 - 0.5x = 0.0875 - 0.5\times 0.00179 = 0.0866 M$

Answer:

$[NO] = 0.00179M$

$[N_2] = 0.0866 M$

$[H_2] = 0.0866 M$