the value of Kc = 4.24 at 800K for the reaction CO(g) + H2O(g) CO2(g) + H2(g) Calculate the equilibrium concentration of CO2 ,H2O ,CO , H2 at 800K ,if only CO and H2O are pesent initially at concentration of 0.10M each. Why didn't we take 0.194 wala value for the calculation
Answers
Answered by
473
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
initial
concentration= 0.1M 0.1M 0 0
concentration
in equilibrium = 0.1-x 0.1-x x x
Hence, equilibrium constant can be written as,
Kc = x^2 /(0.1-x)^2 = 4.24
x^2 = 4.24(0.01 + x^2 -0.2x)
x^2 = 0.0424 + 4.24x^2 - 0.848x
3.24x^2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax + bx + c = 0,
Thus solving we get two values of x
x1 = 0.067 x2 = 0.194
Neglecting x2 = 0.194 because x could not be more than initial concentration.
Hence the equilibrium concentrations are, [CO ] = [H ] = x = 0.067 M
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 - x = 0.1 - 0.067 =0.033 M
i hope it will help you
regards
initial
concentration= 0.1M 0.1M 0 0
concentration
in equilibrium = 0.1-x 0.1-x x x
Hence, equilibrium constant can be written as,
Kc = x^2 /(0.1-x)^2 = 4.24
x^2 = 4.24(0.01 + x^2 -0.2x)
x^2 = 0.0424 + 4.24x^2 - 0.848x
3.24x^2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax + bx + c = 0,
Thus solving we get two values of x
x1 = 0.067 x2 = 0.194
Neglecting x2 = 0.194 because x could not be more than initial concentration.
Hence the equilibrium concentrations are, [CO ] = [H ] = x = 0.067 M
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 - x = 0.1 - 0.067 =0.033 M
i hope it will help you
regards
Answered by
146
For the reaction,
CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g)
Initial concentration:
0.1M0.1 M00
Let x mole per litre of each of the productbe formed.
At equilibrium:
0.1 - x M0.1 - x Mx Mx M
where x is the amount of CO2 and H2 at equilibrium. .
Hence, equilibrium constant can be written as,
Kc= x2/(0.1-x)2 = 4.24
x2 = 4.24(0.01 + x2-0.2x)
x2 = 0.0424 + 4.24x2-0.848x
3.24x2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax2 + bx + c = 0,
Thus solving we get two values of x
x1= 0.067 x2= 0.194
Neglecting x2= 0.194 becuase x could not be more than initial concentration.
Hence the equilibrium concentrations are,
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 – 0.067 = 0.033 M
CO (g) + H2O (g) ⇔ CO2 (g) + H2 (g)
Initial concentration:
0.1M0.1 M00
Let x mole per litre of each of the productbe formed.
At equilibrium:
0.1 - x M0.1 - x Mx Mx M
where x is the amount of CO2 and H2 at equilibrium. .
Hence, equilibrium constant can be written as,
Kc= x2/(0.1-x)2 = 4.24
x2 = 4.24(0.01 + x2-0.2x)
x2 = 0.0424 + 4.24x2-0.848x
3.24x2 – 0.848x + 0.0424 = 0
a = 3.24, b = – 0.848, c = 0.0424
for quadratic equation ax2 + bx + c = 0,
Thus solving we get two values of x
x1= 0.067 x2= 0.194
Neglecting x2= 0.194 becuase x could not be more than initial concentration.
Hence the equilibrium concentrations are,
[CO2] = [H2] = x = 0.067 M
[CO] = [H2O] = 0.1 – 0.067 = 0.033 M
Similar questions
Math,
8 months ago
Biology,
8 months ago
Science,
8 months ago
Science,
1 year ago
Social Sciences,
1 year ago