please help with 21!! The equilibrium constant is 4.31 x 10 4 at 375 degree C: N
ID: 889812 • Letter: P
Question
please help with 21!! The equilibrium constant is 4.31 x 10 4 at 375 degree C: N2(g)*3H2(g) 2NH3(g) In a certain experiment, a chemist starts with 0.862atm N? and 0.373 atm H2 in a reaction vessel at 375 degree C. Calculate the total pressure in the reaction vessel at equilibrium. Nz(g) + 3H2(g) 2N H3(g) 0.8 2 atm Kp = 4.31 times 10 = [2x) 0 -x -3x -2x [0.842 - x] [373-3x] 0.8 2-x 373-3x 2x x = o2121 Total A student decomposes potassium chlorate and collects 36.5 cm3 of oxygen gas over water at 23.0 degree C. The laboratory manometer roads 751 torr. The vapor pressure o water at 23.0 C is 21.1 torr. Find the volume the dry oxygen would occupy at 0 degree C and 1.000 atm. How many grams of potassium chlorate were present before the reaction?Explanation / Answer
Solution :-
Balanced reaction equation
N2(g) + 3H2(g) --------------- > 2 NH3(g)
Kp = 4.31*10^-4
Initial partial pressure of N2 = 0.862 atm and H2 = 0.373 atm
Total pressure at equilibrium = ?
Lets first make ICE table and calculate the equilibrium partial pressures of the each gas
N2(g) + 3H2(g) --------------- > 2 NH3(g)
0.862 0.373 0
-x -3x +2x
0.862-x 0.373-3x 2x
Lets write the kp equation
Kp = [NH3]^2 /[N2][H2]^3
Lets put the values in the formula
4.31*10^-4 = [2x]^2/[0.862-x][0.373-3x]^3
4.31*1.^-4 * [0.862-x][0.373-3x]^3 = 4x^2
Solving this equation we get
X= 0.002136
Therefore now lets calculate the equilibrium partial pressures of the each gas
[N2]eq = 0.862 atm – x = 0.862 atm – 0.002136 atm = 0.8598 atm
[H2] eq = 0.373 atm – 3x = 0.373 atm – (3*0.002136 atm ) = 0.3666 atm
[NH3]eq = 2x = 2*0.002136 atm = 0.004272 atm
Total pressure = pN2 + pH2 +pNH3
= 0.8598 atm + 0.3666 atm + 0.004272 atm
= 1.23 atm
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