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densk [106]
3 years ago
7

Consider the reversible reaction in aqueous medium shown below. At equilibrium, the concentrations of A, B, C, and D are 0.0053

M, 0.0085 M, 0.0040 M, and 0.0035 M, respectively. What is the equilibrium constant for the reaction? A+3B⇌3C+D
Chemistry
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

The equilibrium constant for the reaction is 0.0689

Explanation:

<u>Step 1:</u> Data given

The concentrations of A = 0.0053 M

The concentration of B = 0.0085 M

The concentration of C = 0.0040 M

The concentration of D = 0.0035 M

<u>Step 2:</u> The balanced equation:

A+3B⇌3C+D

<u>Step 3:</u> Calculate the equilibrium constant

K = [C]³[D] / [A][B]³

K = (0.0040)³(0.0035) / (0.0053)(0.0085)³

K = 0.0689

The equilibrium constant for the reaction is 0.0689

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Read 2 more answers
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
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