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Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How?
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Consider an equation x ->2y (in equilibrium) with equilibrium constant...
Equation and Equilibrium Constant
The given equation is x -> 2y and its equilibrium constant is kc=3.6M at 298k.
Initial Concentrations
The initial concentrations are [x]=1.0M and [y]=0M.
Equilibrium Concentration of X
To determine the equilibrium concentration of x, we need to use the equilibrium constant expression:
kc = [y]^2 / [x]
We can rearrange this equation to solve for [x]:
[x] = [y]^2 / kc
Since the initial concentration of y is 0M, we need to use the stoichiometry of the reaction to find the amount of y that will be produced at equilibrium. For every 1 mole of x that is consumed, 2 moles of y will be produced. Therefore, at equilibrium, the concentration of y will be:
[y] = 2[x]
Substituting this expression for [y] into the equation for [x], we get:
[x] = (2[x])^2 / kc
Simplifying this expression, we get:
[x] = 4[x]^2 / kc
Multiplying both sides by kc and rearranging, we get a quadratic equation:
4[x]^2 - kc[x] = 0
Solving for [x] using the quadratic formula, we get:
[x] = kc / (2*2) = 0.40M
Therefore, the equilibrium concentration of x at 298k is 0.40M, which is option (c).
The given equation is x -> 2y and its equilibrium constant is kc=3.6M at 298k.
Initial Concentrations
The initial concentrations are [x]=1.0M and [y]=0M.
Equilibrium Concentration of X
To determine the equilibrium concentration of x, we need to use the equilibrium constant expression:
kc = [y]^2 / [x]
We can rearrange this equation to solve for [x]:
[x] = [y]^2 / kc
Since the initial concentration of y is 0M, we need to use the stoichiometry of the reaction to find the amount of y that will be produced at equilibrium. For every 1 mole of x that is consumed, 2 moles of y will be produced. Therefore, at equilibrium, the concentration of y will be:
[y] = 2[x]
Substituting this expression for [y] into the equation for [x], we get:
[x] = (2[x])^2 / kc
Simplifying this expression, we get:
[x] = 4[x]^2 / kc
Multiplying both sides by kc and rearranging, we get a quadratic equation:
4[x]^2 - kc[x] = 0
Solving for [x] using the quadratic formula, we get:
[x] = kc / (2*2) = 0.40M
Therefore, the equilibrium concentration of x at 298k is 0.40M, which is option (c).
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Consider an equation x ->2y (in equilibrium) with equilibrium constant...
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Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How?
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Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How?.
Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How?.
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Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How?, a detailed solution for Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? has been provided alongside types of Consider an equation x ->2y (in equilibrium) with equilibrium constant kc=3.6M at 298k. If the initial concentration are [x]=1.0M and [y]=0M. The equilibrium concentration x at 298k is _? a)0.33M b)0.36M c )0.40M. d) 0.60M Correct option is c. How? theory, EduRev gives you an
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