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A researcher is studying the decomposition of A as shown by the general reaction below:2 A(g) 2 B(g) + 2 C(g)Initially, the scientist fills an evacuated 4.956 L flask with 9.570 x 10-1 moles of species A. Upon equilibrium, it is determined that the concentration of A is 1.724 x 10-1 M. Calculate Kc.

Question

A researcher is studying the decomposition of A as shown by the general reaction below:

2 A(g) 2 B(g) + 2 C(g)

Initially, the scientist fills an evacuated 4.956 L flask with 9.570 x 10-1 moles of species A. Upon equilibrium, it is determined that the concentration of A is 1.724 x 10-1 M. Calculate Kc.

 

check_circleAnswer
Step 1

Kc of the reaction is calculted using ICE table.

An equilibrium constant (K) is the ratio of concentration of products and reactants raised
to appropriate stoichiometric coefficient at equlibrium
For the general reaction
aА + bв — сС+dD
The relative strength of reaction can be also expressed quantitatively with an equilibrium
constant as follows:
(aC) (aD)
К.
(aA)
(1)
aB
An equilibrium constant is (K ).
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An equilibrium constant (K) is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium For the general reaction aА + bв — сС+dD The relative strength of reaction can be also expressed quantitatively with an equilibrium constant as follows: (aC) (aD) К. (aA) (1) aB An equilibrium constant is (K ).

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Step 2

concentration of A is calculated.

The reaction is shown below
2A (g)2B (g)
2C (g)
K of the reaction is calculated as follows,
Moles
Concentration Volume
=
Concentration ofA = Moles
Volume
Concentration of A = 9.570 x10
4.956
Concentration of A=0.193
help_outline

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The reaction is shown below 2A (g)2B (g) 2C (g) K of the reaction is calculated as follows, Moles Concentration Volume = Concentration ofA = Moles Volume Concentration of A = 9.570 x10 4.956 Concentration of A=0.193

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Step 3

At equilibrium the concentration of...

ICE table
2C (g)
2B (g
2A(g)
0.193 M
Initial
0
0
concentration
+2x
Change
-2x
+2x
At equilibrium
0.193-2x
2x
2x
According to the equation,
0.193-2x-1.724 x1 0
2x 0.0206
x0.0103
help_outline

Image Transcriptionclose

ICE table 2C (g) 2B (g 2A(g) 0.193 M Initial 0 0 concentration +2x Change -2x +2x At equilibrium 0.193-2x 2x 2x According to the equation, 0.193-2x-1.724 x1 0 2x 0.0206 x0.0103

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