The following reaction was performed in a sealed vessel at 772 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.90M and [I2]=2.95M . The equilibrium concentration of I2 is 0.0500 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

The value of for the given equation is 160.2

Explanation:

We are given:

Initial concentration of hydrogen gas = 3.90 M

Initial concentration of iodine gas = 2.95 M

Equilibrium concentration of iodine gas = 0.0500 M

The given chemical equation follows:

Initial:          3.90     2.95

At eqllm:   3.90-x    2.95-x      2x

Evaluating the value of 'x':

So, equilibrium concentration of hydrogen gas = (3.95 - x) = (3.95 - 2.9) = 1.05 M

Equilibrium concentration of HI gas = x = 2.9 M

The expression of for above equation follows:

Putting values in above equation, we get:

Hence, the value of for the given equation is 160.2

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