Q 25 equilibrium Q25 In a 7 0 l, evacuated chamber, 0 50 mol H2 and 0 50 - Chemistry - Equilibrium

Question

Q 25 equilibrium

Q25 In a 7 0 l, evacuated chamber, 0 50 mol H2 and 0 50
mol l2 react at 427 ° C

H2(g) + l2(g) ⇌ 2HI (g) At the given
temperature, Kc= 49 for the reaction

(i) What is the total pressure (atm) in the chamber?

(A) 83 14 (B) 831 4 (C) 8 21 (D) None

(ii) What is the value of Kp?

(A) 7 (B) 49 (C) 24 5 (D) None

(iii) How many moles of the iodine remain unreacted at equilibrium
?

(A) 0 388 (B) 0 112 (C) 0 25 (D) 0 125

(iv) What is the partial pressure (atm ) of Hl in the equilibrium
mixture ?

(A) 6 385 (B) 12 77 (C) 40 768 (D) 646 58

Yasodharan · Accepted Answer

Dear Student,

The computation are as expressed below,
Q no:25(i) P=nRTV=(0 5+0 5)(0 0821)(427+273)7=8
21 atm(ii) Kp=Kc(RT)∆n ⇒Kp=Kc  {∆n = 2-(1+10 =0}     Kp=49(iii)   H2   +    I2     ⇔    2HI         0
5        0
5              0         -x        -x            2x       0
5-x     0
5-x       2xK=[HI]2[H2][I2]⇒49=(2x)2(0
5-x)(0 5-x)=(2x)2(0
5-x)2taking square root we get,49=(2x)2(0
5-x)27 = 2x0 5-x⇒3 5-7x = 2x3
5=9xx = 0
39so moles of unreacted I2remained at equilibrium = 0
5-0 39=0
11 mol(iv) pH2=mol of H2Total mol×total pressure=0
11×8 211=0
9031 atm      pH2=pI2=0
9031 atm      pHI=Total pressure -(pH2+pI2) =8
21-(0 9031+0
9031)              = 6
4038 atm

Answers
(i) Option (C) 8 21
(ii) Option (B) 49
(iii) Option (A) 0 388
(iv) Option (A) 6 385

Regards