KC for reaction is 55.0 at 700C calculate the equilibrium concentration of reaction and products when the initial consultation of h2 and i2 are 2.55
To calculate the equilibrium concentrations of the reaction and products, we need to use the equation for the equilibrium constant (Kc) and the given initial concentrations.
The balanced equation for the reaction is:
H₂(g) + I₂(g) ⇌ 2HI(g)
We are given the value of Kc, which is 55.0. This means that:
Kc = [HI]² / ([H₂] * [I₂])
We are also given the initial concentrations of H₂ and I₂, which are both 2.55 M.
Now, let's assume that at equilibrium, the concentration of HI is "x" M. Since 2 moles of HI are formed for every mole of H₂ and I₂, the equilibrium concentrations can be written as:
[H₂] = 2.55 - x
[I₂] = 2.55 - x
[HI] = x
Substituting these values into the equation for Kc, we can solve for "x":
55.0 = (x)² / ((2.55 - x) * (2.55 - x))
Simplifying this equation gives us a quadratic equation:
55.0 * (2.55 - x)² = (x)²
55.0 * (6.5025 - 5.1x + x²) = x²
357.6375 - 280.5x + 55x² = x²
54x² - 280.5x + 357.6375 = 0
Solving this quadratic equation will give us the value of "x", which represents the equilibrium concentration of HI.