At a particular temperature a 2.00-L flask at equilibrium contains 2.80 ✕ 10-4 mol N2, 2.50 ✕ 10-5 mol O2, and 2.00 ✕ 10-2 mol N2O. Calculate K at this temperature for the following reaction. 2 N2(g) + O2(g) equilibrium reaction arrow 2 N2O(g)

Answer : The value of equilibrium constant (K) is,

Explanation :

First we have to calculate the concentration of

and,

and,

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

Thus, the value of equilibrium constant (K) is,

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

you need to have a picture of the atom that way we can answer it