At 450°c, ammonia gas will decompose according to the following equation: 2 nh3 (g)  n2 (g) + 3 h2 (g) kc = 4.50 at 475˚c an unknown quantity of nh3 is placed in a reaction flask (with no n2 or h2) and is allowed to come to equilibrium at 475°c. the equilibrium concentration of h2 is then determined to be 0.252 m. determine the initial concentration of nh3 placed in the flask

0.2024 M

Explanation:

For the decomposition reactio given, let's do an equilibrium chart. Let's call the initial concentration of NH₃ as C:

2NH₃(g) ⇄ N₂(g) + 3H₂(g)

C 0 0 Initial

-2x +x +3x Reacts (stoichiometry is 1:1:3)

C - 2x x 3x Equilibrium

3x = 0.252

x = 0.084 M

The equilibrium constant (Kc) is the multiplication of the concentrations of the products elevated by their coefficients, divided by the multiplication of reactants concentrations elevated by their coefficients.

Kc = ([H₂]³*[N₂])/([NH₃]²)

4.50 = [(0.252)³*(0.084)]/(C - 2*0.084)²

4.50 = 0.00533/(C - 0.168)²

4.50 = 0.00533/(C² - 0.336C + 0.028224)

4.50C² - 1.512C + 0.127008 = 0.00533

4.50C² - 1.512C + 0.121678 = 0

Solving the equation by a graphic calculator, for C > 0.168

C = 0.2024 M

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

i think is conclusions.i think so