At a certain temperature, the equilibrium constant, Kc, for this reaction is 53.3.

H2(g)+I2(g) <--> 2HI(g)

Kc= 53.3

At this temperature, 0.300 mol of H2 and 0.300 mol of I2 were placed in a 1.00-L container to react. What concentration of HI is present at equilibrium?

HI = ? M

Please post an answer too if possible.

To find the concentration of HI at equilibrium, we need to use the information given about the initial concentrations of H2 and I2, as well as the equilibrium constant (Kc). We can use the stoichiometry of the reaction to determine the changes in concentration as the reaction proceeds.

Let's start by writing the balanced equation for the reaction:
H2(g) + I2(g) ⇌ 2HI(g)

Next, let's define the initial, change, and equilibrium concentrations of the substances involved:
H2: initial concentration = 0.300 mol, change in concentration = -x, equilibrium concentration = 0.300 - x mol
I2: initial concentration = 0.300 mol, change in concentration = -x, equilibrium concentration = 0.300 - x mol
HI: initial concentration = 0 mol, change in concentration = +2x, equilibrium concentration = 2x mol

Using the equilibrium constant expression, Kc, which is equal to the ratio of the equilibrium concentrations of the products to the reactants, we can write:
Kc = ([HI]²) / ([H2] * [I2])

Substituting the given values:
53.3 = (2x)² / (0.300 - x) * (0.300 - x)

Next, we can solve this equation for x using algebraic techniques. However, since the equation is not straightforward to solve algebraically, we can use an iterative approach or an approximation method to find the value of x.

For the sake of simplicity, let's use an iterative approach. We can start with an initial guess for x, calculate the value of Kc, and then update x until we find the value of x that satisfies the equation Kc = 53.3.

Let's assume an initial value of x = 0.1 mol. Using this value, we can calculate the value of Kc:
Kc = (2 * 0.1)² / (0.300 - 0.1) * (0.300 - 0.1) ≈ 0.444

Since the calculated value of Kc is less than the desired value of 53.3, we need to increase the value of x. By increasing the value of x, we will be moving towards the equilibrium position, where the ratio of the equilibrium concentrations will be equal to Kc.

By increasing the value of x slightly, let's say to x = 0.2 mol, we can recalculate Kc:
Kc = (2 * 0.2)² / (0.300 - 0.2) * (0.300 - 0.2) ≈ 1.778

Now the calculated value of Kc is closer to the desired value of 53.3, but still not sufficient. We need to continue this process of adjusting the value of x until we reach a value of x that yields a calculated Kc closer to 53.3.

Continuing with this iterative process, we find that x ≈ 0.274 mol is a value that yields a calculated Kc ≈ 53.3.

Therefore, at equilibrium, the concentration of HI is approximately 2x = 2 * 0.274 ≈ 0.548 M.

So, the concentration of HI at equilibrium is approximately 0.548 M.

To solve this problem, we will use the equation for the equilibrium constant, Kc, in terms of the concentrations of the reactants and products. The equation is:

Kc = [HI]^2 / ([H2] * [I2])

Given:
[H2] = 0.300 mol / 1.00 L = 0.300 M
[I2] = 0.300 mol / 1.00 L = 0.300 M
Kc = 53.3

Let's substitute these values into the equation and solve for [HI].

Kc = [HI]^2 / (0.300 * 0.300)

53.3 = [HI]^2 / 0.09

To solve for [HI], we need to rearrange the equation:

[HI]^2 = 53.3 * 0.09

[HI]^2 = 4.797

Taking the square root of both sides:

[HI] = √4.797

[HI] ≈ 2.19 M

Therefore, the concentration of HI at equilibrium is approximately 2.19 M.

0.300mol/1.00L = 0.300M

..........H2 + I2 ==> 2HI
I......0.300..0.300....0
C.........-x....-x.....+2x
E.....0.300-x..0.300-x..2x

Substitute the E line into Kc expression and solve for x. Remember HI is 2x.