Date Advanced Equilibrium and Le Chatelier\'s Principle Free Response Question M

Question

Date Advanced Equilibrium and Le Chatelier's Principle Free Response Question Make up and Retake 1. Given the following equilibrium reaction Kp-0.052 at 27.0°C N2H: (g) N2 (g)+H:(g) a. Ir0.30 mol of NaH is placed into an empty 4.00 L flask, what will be the partial pressures of all the gases at equilibrium? atm b. For the same reaction, if the flask was instead charged with 0.60 atm NaHh, 0.60 atm No, and 4.0 of Hh, would the partial pressure increase for any of the gases that we started with afer this misture reached equilibrium? Justify your answer.

Explanation / Answer

For the reaction N2H2(g) <------>N2(g)+ H2(g)

Moles of N2H2(n)=0.3, V= volume = 4L. R= gas constant = 0.0821 L.atm/mole.K, T= 27 deg.c= 27+273= 300K

Pressure of N2H4, PN2H2= nRT/V= 0.3*0.0821*300/4=1.85 atm

Q= reaction coefficient = PN2*PH2/PN2H2= 0/1.85=0<K, so the reaction proceeds forward so as to make, Q=K at equilibrium

Let P= drop in pressure of N2O4 to reach equilibrium. Hence at Equilibrium

PN2H2= 1.85-P, PN2=PH2=P

K= PN2*PH2/PN2H2= P2/(1.85-P)=0.052

Solving for P using excel, P=0.0285, so at equilibrium, PN2H4= 1.85-0.0285= 1.565 atm, PN2=PH2=0.0285 atm

2. When PN2H2= 0.6 atm, PN2=0.6 atm, PH2=4

Q= 4*0.6/0.6=4 atm>K, reaction proceeds backward so as to reduce Q.

Let P = rise in pressure of N2H2. So at equilibrium, PN2H2=0.6+P, PN2=0.6-P and PH2=4-P

KP= (0.6-P)*(4-P)/(0.6+P)=0.052, when solved using excel, P=0.5847 atm

So at Equilibrium, PN2=0.6-0.5847= 0.0153 atm, PH2=4-0.5847=3.4153 atm and PN2H2=0.6+0.5847=1.1847 atm

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.