Could I get a step by step explanation and answer to 3 and 4 please? 3. Nitrogen
ID: 1025195 • Letter: C
Question
Could I get a step by step explanation and answer to 3 and 4 please?
3. Nitrogen trifluoride decomposes to form nitrogen and fluorine gases according to the following equation: 2NFdg)- N2(g) + 3F2(g) When 2.06 mol of NFs is placed in a 2.00-L container and allowed to come to equilibrium at 800 K, the mixture is found to contain 0.0227 mol of N2. What is the value of K at this temperature? a) 1.77 × 10-6 b) 4.43 x 10-7 c) 1.91 x 103 d) 1.83 x 103 e) 1.73 x 106 4. At 400 K, an equilibrium mixture of H2, 12, and HI consists of 0.054 mol Ha, 0.019 mol Ib, and 0.059 mol HI in a 1.00-L flask. What is the value of Ke for the following equilibrium? (R-0.0821 L-atm/(K·mol)) 2HI(g) H2(g)+ 12(g) a) 3.4 b) 21 c) 0.29 d) 0.017 e) 58
Explanation / Answer
[NF3] = no of moles of NF3 /volume in L
= 2.06/2 = 1.03M
[N2] = no of moles of NF3 /volume in L
= 0.0227/2 = 0.01135M
2NF3(g) -------------> N2(g) + 3F2(g)
I 1.03 0 0
C -2*0.01135 0.01135 3*0.01135
E 1.0073 0.01135 0.03405
Kc = [N2][F2]^3/[NF3]^2
= 0.01135*(0.03405)^3/(1.0073)^2
= 4.4*10^-7
Dn = (1+3)-2 = 2
Kp = Kc(RT)^Dn
Kp = 4.42*10^-7 (0.0821*800)^2 = 1.91*10^-3
c. 1.91*10^-3
2HI(g) ---------> H2(g) + I2(g)
[H2] =0.054mole/1L = 0.054M
[I2] = 0.019/1 = 0.019M
[HI] = 0.059/1 = 0.059M
Kc = [H2][I2]/[HI]^2
= 0.054*0.019/(0.059)^2
= 0.29
Dn = (1+1)-2 = 0
Kp = Kc(RT)^Dn
= 0.29(0.0821*400)^0
= 0.29
c. 0.29