For the reaction 2HI=H2+I2.one mole of HI is introduced into a vessel at constant temperature.calculate the no of moles of I2 when equilibrium is reached (Kc=0.0156)
Plz help me and give complete solution
I can't understand .312x how this value come i mean from where?
To find the number of moles of I2 when equilibrium is reached, we first need to calculate the concentration of HI at equilibrium using the given value of Kc.
The balanced chemical equation for the reaction is:
2HI = H2 + I2
Let "x" moles of HI react to form x moles of H2 and I2. Therefore, at equilibrium, the concentrations are:
[HI] = (1 - x) moles/L
[H2] = x moles/L
[I2] = x moles/L
Using the expression for Kc, we can write:
Kc = [H2][I2] / [HI]^2
Substituting the concentrations into the expression:
0.0156 = (x)(x) / (1 - x)^2
0.0156 = x^2 / (1 - x)^2
To solve this equation, we can rearrange it as follows:
0.0156(1 - x)^2 = x^2
Expand the equation:
0.0156(1 - 2x + x^2) = x^2
0.0156 - 0.0312x + 0.0156x^2 = x^2
Rearrange the equation by bringing all terms to one side:
0.0156x^2 + 0.0312x - 0.0156 = 0
Now, this is a quadratic equation. You can solve it using the quadratic formula or factoring. I'll show you how to solve it using the quadratic formula:
The quadratic formula states that for an equation ax^2 + bx + c = 0, the roots (solutions) are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 0.0156, b = 0.0312, and c = -0.0156. Substituting these values into the formula:
x = (-(0.0312) ± √((0.0312)^2 - 4(0.0156)(-0.0156))) / (2(0.0156))
Simplifying the equation gives two possible values for x. Remember, x represents the number of moles of HI that reacted. Therefore, we consider only the positive value for x since we can't have a negative number of moles:
x = 0.556 moles (approximately)
Since the balanced equation shows that 2 moles of HI react to form 1 mole of I2, the number of moles of I2 at equilibrium is half the value of x:
Number of moles of I2 = 0.556 / 2 = 0.278 moles
K=x^2/(1-(2x)) = 0.0156
It comes from 0.0156 * 2
Right
2HI>>H2 +I2
letting x be the conc of (I2)
The initial concentration of HI is 1mole/vol
K=x^2/(1-(2x))
.0156-.312x-x^2=0
x=(.312+-sqrt(.312^2+.0624))/(-2)
x= .0438 moles
check my math, it is easy to make an error.