In: Chemistry
A) For the reaction: 3A(g)+2B(g)⇌C(g) Kc = 63.4 at a temperature of 377 ∘C .
Calculate the value of Kp. Express your answer numerically.
B) For the reaction X(g)+3Y(g)⇌3Z(g) Kp = 2.91×10−2 at a temperature of 229 ∘C .
Calculate the value of Kc. Express your answer numerically.
C) For the reaction 2CH4(g)⇌C2H2(g)+3H2(g) K = 0.155 at 1774 ∘C .
What is Kp for the reaction at this temperature? Express your answer numerically.
D) For the reaction N2(g)+3H2(g)⇌2NH3(g) Kp = 4.40×10−3 at 271 ∘C .
What is K for the reaction at this temperature? Enter your answer numerically.
E) The following reaction was performed in a sealed vessel at 772 ∘C : H2(g)+I2(g)⇌2HI(g)
Initially, only H2 and I2 were present at concentrations of [H2]=3.00M and [I2]=2.80M. The equilibrium concentration of I2 is 0.0400 M . What is the equilibrium constant, Kc, for the reaction at this temperature?
Express your answer numerically.
Thank you so much in advance!
we know that
Kp = Kc (RT)^dn
dn = moles of gases in products - moles of gases in reactants
A)
given
3A + 2B --> C
dn = 1 - 2 - 3 = -4
so
Kp = Kc (RT)^-4
so
Kp = 63.4 x (0.0821 x 650)^-4
Kp = 7.82 x 10-6
B)
given
X + 3Y ---> 3Z
dn = 3 - 3 - 1
dn = -1
so
Kp = Kc (RT)^-1
Kc = Kp (RT)
so
Kc = 2.91 x 10-2 x (0.0821 x 502)
Kc = 1.2
C)
given
2CH4 ---> C2H2 + 3H2
dn = 1 + 3 - 2 = 2
so
Kp = Kc (RT)^2
so
Kp = 0.155 x (0.0821 x 2047)^2
Kp = 4378
D)
given
N2 + 3H2 ---> 2NH3
dn = 2 - 1 - 3 = -2
so
Kp = Kc (RT)^-2
Kc = Kp (RT)^2
Kc = 4.4 x 10-3 ( 0.0821 x 544)^2
Kc = 8.77
E)
H2 + I2 --> 2HI
using ICE table
at equilibrium
[H2] = 3 - x
[I2] = 2.8 - x
[HI] = 2x
given
[I2] = 0.04
so
2.8 - x = 0.04
x = 2.76
so
[H2] = 3 - x = 3 - 2.76 = 0.24
[I2] = 2.8 - x = 0.04
[HI] = 2x = 2 * 2.76 = 5.52
now
Kc = [HI]^2 / [H2] [I2]
Kc = [5.52]^2 / [0.24] [0.04]
Kc = 3174
so
the value of equilibrium constant is 3174