(the fraction of the iodine that is dissociated) was 0.0274 Calculate K and Kp a
ID: 555731 • Letter: #
Question
(the fraction of the iodine that is dissociated) was 0.0274 Calculate K and Kp at that temperature. 4.6. It has bcen observed with the ammonia equilibrium PROBLEMS Equilibrium Constants 4.1. A reaction occurs according to the equation N2 + 3H2 2NH3 If in a volume of 5 dm we start with 4 mol of pure A and find that 1 mol of A remains at equilibrium, what is the equilibrium constant K? 4.2. The equilibrium constant for a reaction that under certain conditions the addition of nitrogen to an equilibrium mixture, with the temperature and pressure held constant, causes further dissociation of ammonia. Explain how this is possible. Under what particular conditions would you expect this to occur is 0.1. What amount of A must be mixed with 3 mol of B to Would it be possible for added hydrogen to produce the same effect? yield, at equilibrium, 2 mol of Y? 4.3. The equilibrium constant for the reaction 4.7. Nitrogen dioxide, NO2, exists in equilibrium wit dinitrogen tetroxide, N,O N 204(g) 2NO2(g) is 0.25 dm mol 2. In a volume of 5 dm', what amount of A must be mixed with 4 mol of B to yield I mol of Z at equilibrium? 4.4. The equilibrium constant K, for the reaction At 25.0 °C and a pressure of 0.597 bar the density of gas is 1.477 g dm. Calculate the degree of dissocia under those conditions, and the equilibrium constants Kp, and K What shift in equilibrium would occur i pressure were increased by the addition of helium gas? is 0.0271 mol dm at 1100 K. Calculat Kp at that48. At 25.0°C the equilibrium ture. 4.5. When gaseous iodine is heated, dissociation occurs: 2NOBr(g) 2NO(g) + Br2(g) It was found that when 0.0061 mol of iodine was placed in a volume of 0,5 dm' at 900 K, the degree of dissociation is rapidly established. When 1.10 g of NOBr is pres Calculate the equilibrium constants Kp. K, and K, a 10-dm3 vessel at 2 25.0 °C the pressure is 0.35Explanation / Answer
concentration of Iodine = moles of iodine/volume
1dm3= 1L, 0.5dm3= 0.5L
concentration of iodine = 0.0061/0.5=0.0122M
the reaction is I2 <---->2I(g)
let x= drop in concentration of I2. At equilibrium [I2]= 0.0122-x and [I]= 2x
fraction dissociated = x/0.0122= 0.0274, x= 0.0122*0.0274 =0.000334
at equilibrium, [I2] =0.0122-0.000334 = 0.011866 , [I]= 2*0.000334 =0.000668
KC= [I]2/ [I2] =0.000668*0.000668/0.011866 = 3.76*10-5
Kp = KC*(RT)deltan , deltan= mole of products- moles of reactants= 2-1=1
KP= 3.76*10-5 *(0.0821 L*900).atm/mole.K =0.002778
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.