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.
Plz help me and give complete solution
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.
It comes from 0.0156 * 2
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