Question #19411

Hydrogen, iodine, and hydrogen iodide gases have reached equilibrium in a container at 700K. Equilibrium partial pressures are 0.28 atm HI, 0.033 atm H2, and 0.045 atm I2. For H2 (g) + I2 (g) --> 2HI, K=53. Half the HI is removed form the container and equilibrium is reestablished. Calculate the new partial pressure (in atm) of HI in the new equilibrium mixture.

Expert's answer

K = [HI]^{2}/[I_{2}][H_{2}] = P^{2}(HI)/P(I_{2})*P(H_{2}) = 53

half of HI is proportional to pressure = 0.28/2 = 0.14

53 = (0.14+x^{2} )^{2} / (0.033-x)*(0.045-x)

x=....

new p of HI = 0.14+x (from upper equation)

half of HI is proportional to pressure = 0.28/2 = 0.14

53 = (0.14+x

x=....

new p of HI = 0.14+x (from upper equation)

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