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hodyreva [135]
3 years ago
12

What will happen if pressure is increased?

Chemistry
2 answers:
fgiga [73]3 years ago
8 0

When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.

melomori [17]3 years ago
8 0
Increase the pressure on what ?
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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

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3 years ago
Fluorine has 7 valence electrons what type of bond is likely to form between two atoms?
Semmy [17]
I'm not quiet sure...possibly an ionic bond.
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How many molecules of CO2 are in 8.6 mol of CO2?
Anastasy [175]

Answer: You’d have to convert from mol to molecule.

Explanation: Like this:

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A box has a weight of 120 lbs and the bottom of the box is 12 in2 . What is the pressure the box exerts on the floor?
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How can I help with this question?
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Roman mixes 12 liters of 8% acid solution with a 20% acid solution, which results in a 16% acid solution. Find the number of lit
inysia [295]
Let the 8% solution be A, the 20% solution be B and the final solution be C.

C = A + B
C = 12 + B

0.16C = 0.08(12) + 0.2(B)
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0.96 = 0.04B
B = 24 Liters

C = 12 + 24
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3 0
3 years ago
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