Given K = 3.45 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g)?

[匿名]

2019-02-23 3:10 pm

and K = 7.04 at 45°C for the reaction
2 A(g) + D(g) equilibrium reaction arrow C(g)

a) what is the value of K at the temperature for the following reaction?
C(g) + D(g) equilibrium reaction arrow 2 B(g)
_________

b) What is the value of Kp at 45°C for the same reaction?

_________

c) Starting with 1.57 atm partial pressures of both C and D, what is the mole fraction of B once equilibrium is reached?
__________

更新1:

Edit: I got a and b. They're 0.5914, but still need help on getting c.

回答 (1)

胡雪8°

2019-02-24 1:10 am

✔ 最佳答案

For a) and b)
K = Kp = 7.04/3.45² = 0.591 (to 3 sig. fig.)
The answer is to 3 sig. fig. because the equilibrium constants for the two given equilibria are to 3 sig. fig.

c)
__________________ C(g) __+ __ D(g) __ ⇌ __ 2B(g) ___ K = 0.591
Initial (atm): _______ 1.57 ______ 1.57 ________ 0
Change (atm): ______ -y ________ -y _________ +2y
Equilibrium (atm): _ 1.57 - y ___ 1.57 - y _______ 2y

At equilibrium: K = [B]² / ([C] [D])
0.591 = (2y)² / (1.57 - y)²
√0.591 = 2y / (1.57 - y)
1.57√0.591 - y√0.591 = 2y
2y + y√0.591 = 1.57√0.591
y(2 + √0.591) = 1.57√0.591
y = 1.57√0.591 / (2 + √0.591)
y = 0.436 (to 3 sig, fig.)

Mole fraction of B at equilibrium
= [B] / (Sum of concentration)
= 2×0.436 / (1.57 - 0.436 + 1.57 - 0.436 + 2×0.436)
= 0.278 (to 3 sig. fig.)

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