The pressure of iodine gas at a particular temperature is found to be `0.111atm`, where as the expected pressure is `0.074 atm`, the increased pressure is due to `I_(2)hArr 2I`. Calculate `K_(p)` for this equilibrium.

The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation I_(2) hArr 2I . Calculate K_(p) .

The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation I_(2) hArr 2I . Calculate K_(p) .

At 1000^@C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation, I_2 hArr 2I . Calculate K_p . Also find out pressure at which I_2 will be 90% dissociation at 1000^@C .

At 1000^@C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation, I_2 hArr 2I . Calculate K_p . Also find out pressure at which I_2 will be 90% dissociation at 1000^@C .

At a certain temperature & total pressure of 10^(5)Pa , iodine vapour contians 40% by volume 1 atoms, I_(2)(g)hArr 2I(g) . Calculate K_(p) for the equilibrium.

At a certain temperature and total pressure of 10^(5) Pa, iodine vapours contains 40% by volume of 1 atoms. I_(2)(g) hArr 2 I(g) Calculate K_(p) for the equilibrium.

At a certain temperature and total pressure of 10^5 Pa,iodine vapour contains 40% by volume of 1 atoms I_2(g) hArr 2I(g) Calculate K_p for the equilibrium.