Respuesta :
Answer: The equilibrium concentration of A, B and C are 0.02 M, 2.38 M and 3.04 M respectively.
Explanation:
We are given:
Initial moles of A = 2.30 moles
Initial moles of B = 3.90 moles
Volume of the container = 1.00 L
Concentration of a substance is calculated by:
[tex]\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}[/tex]
So, concentration of A = [tex]\frac{2.30}{1.00}=2.30M[/tex]
Concentration of B = [tex]\frac{3.90}{1.00}=3.90M[/tex]
The given chemical equation follows:
[tex]3A(g)+2B(g)\rightleftharpoons 4C(g)[/tex]
Initial: 2.30 3.90
At eqllm: 2.30-3x 3.90-2x 4x
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[C]^4}{[A]^3\times [B]^2}[/tex]
We are given:
[tex]K_c=2.93\times 10^{29}[/tex]
Putting values in above expression, we get:
[tex]2.93\times 10^{29}=\frac{(4x)^4}{(2.30-3x)^3\times (3.90\times 2x)^2}\\\\x=0.76[/tex]
So, the equilibrium concentration of A = [tex](2.30-3x)=(2.30-(3\times 0.76))=0.02M[/tex]
The equilibrium concentration of B = [tex](3.90-2x)=(3.90-(2\times 0.76))=2.38M[/tex]
The equilibrium concentration of C = [tex]4x=(4\times 0.76)=3.04M[/tex]
Hence, the equilibrium concentration of A, B and C are 0.02 M, 2.38 M and 3.04 M respectively.
The equilibrium is the condition of constant rate of reaction. The equilibrium concentration of A is 0.02 M, B is 2.38, and C is 3.04 M.
What is equilibrium constant?
The equilibrium constant is the ratio of the concentration of product and reactant at equilibrium.
The initial concentration of A and B in 1 L of the solution is given as:
[tex]\rm Molarity=\dfrac{Moles}{Volume}[/tex]
The volume is 1L, thus molarity is equal to moles.
So, the concentration of A is 2.30 M, and the concentration of B is 3.90 M.
The equilibrium concentration of A, B, and C is given in the image attached.
The equilibrium constant ([tex]K_c[/tex]) for the reaction is given as:
[tex]K_c=\rm \dfrac{[C]^4}{[A]^3[B]^2}[/tex]
Substituting the value of A, B, C and equilibrium constant:
[tex]2.93\;\times\;10^{29}=\dfrac{(4x)^4}{(2.30-3x)^3\;\times\;(3.90-2x)^2}\\ x=0.76[/tex]
The equilibrium concentration for the reaction is given as:
- [tex]\text A=2.30-3x\\\text A=2.30-3(0.76)\\\text A=0.02\;\text M[/tex]
The concentration of A at equilibrium is 0.02M.
- [tex]\text B=3.90-2x\\\text B=3.90-2(0.76)\\\text B=2.38\;\rm M[/tex]
The concentration of B at equilibrium is 2.38 M.
- [tex]\text C=4x\\\text C=4(0.76)\\\rm C=3.04\;M[/tex]
The concentration of C at equilibrium is 3.04 M.
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