Question

At a certain temperature only 50% HI is dissociated at equilibrium in the reaction 2HI(g)<—-> H2(g)+I2(g) the equilibrium constant for reaction is
1
0.5
0.25

Answer 1

Let the entire reaction take place in a vessel of Volume V Given the equation 2HI= I2 + H2 Here let initial moles of HI=N. Moles vanished=y So at time=0   N=   0       0 At time =t N-2y= y      y Given 50% of HI dissociates. So conc. of HI left=50% So we can say N-2y=50%of N N-2y=N/2 2y=N-N/2 y=N/4 [I2] =N/4V=[H2] [HI] =N/2V So  Kc=[I2].[H2]/[HI]^2 Kc=(N^2/16)/(N^2/4)     =(N^2x4)/(N^2x16)     =4/16     =0.25 So 0.25 is option (c). Is that your answer?

Answer 2

Simply right their initial conc and equilibrium conc and use law of mass action.