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Answer:
0.169
Explanation:
Let's consider the following reaction.
A(g) + 2B(g) ⇄ C(g) + D(g)
We can find the pressures at equilibrium using an ICE chart.
A(g) + 2 B(g) ⇄ C(g) + D(g)
I 1.00 1.00 0 0
C -x -2x +x +x
E 1.00-x 1.00-2x x x
The pressure at equilibrium of C is 0.211 atm, so x = 0.211.
The pressures at equilibrium are:
pA = 1.00-x = 1.00-0.211 = 0.789 atm
pB = 1.00-2x = 1.00-2(0.211) = 0.578 atm
pC = x = 0.211 atm
pD = x = 0.211 atm
The pressure equilibrium constant (Kp) is:
Kp = pC × pD / pA × pB²
Kp = 0.211 × 0.211 / 0.789 × 0.578²
Kp = 0.169
Answer: Temperature in constant pressure is 286 K
Explanation: If pressure remains constant, then V/T = constant.
V1 = 350 ml and T1 = 200 K and V2 = 500 ml. V1/T1 = V2/T2
and T2 = T1· V2 / V1 = 200 K · 500 ml / 300 ml = 285,7 K
Answer:
According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of
and keep same equilibrium constant
Explanation:
In this buffer following equilibrium exists -

So,
is involved in the above equilibrium.
When a strong base is added to this buffer, then concentration of
increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of
and keep same equilibrium constant.
Therefore excess amount of
combines with
to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.