For the following reaction
2 R (g) --> Q (g)
if the equilibrium concentration of R is 0.776 M and that of Q is 1.22 M at 25 ?C, what is the value of the equilibrium constant at this temperature?

a. 0.521
b. 0.494
c. 2.03
d. 1.57

(C) 2.03

Explanation:

Kc (equilibrium constant) = [Q]/[R]^2

[Q] = 1.22 M

[R] = 0.776 M

Kc = 1.22/0.776^2 = 1.22/0.602176 = 2.03 (to 2 decimal places)

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