The reaction below has an equilibrium constant kp=2.2×106 at 298 k. 2cof2(g)⇌co2(g)+cf4(g) you may want to reference (pages 680 - 684) section 15.3 while completing this problem. part a calculate kp for the reaction below. 4cof2(g)⇌2co2(g)+2cf4(g)

Kp is the equilibrium pressure constant calculated from the partial pressures of a reaction equation.
 Kp =[pCF4]*[p CO2] / [p COF2]^2 = 2.2 x 10^6
When the mole fraction gets doubled we have
 Kp = [pCO2]^2*[pCF4]^2 / [pCOF2]^4
 Kp = [[pCF4]*[p CO2] / [p COF2]^2] * 2
Kp = (2.2 * 10^6) * 2
Kp = 4.8 * 10^12

Ans. : frostbite is when exposure to cold temperatures causes freezing to the skin or other tissues. it most commonly affects the hands, feet, and face. the longer areas are exposed to cold, typically the worse the frostbite. the water temperature used for rewarming a frostbitten part should be : (102 -105) f degrees.