At a certain temperature 0.1moles of hydrogen gas and 0.1 moles of iodine gas were placed in 1 litre flask. It was found that the concentration of iodine decreased to 0.02moles per litre. Calculate the value of the Kc for the reaction
initial (H2) = 0.1mol/1 L = 0.1M
initial (I2) = 0.lmol/1 L = 0.1M
......H2 + I2 ==> 2HI
I....0.1...0.1.....0
C....-x....-x......2x
E...0.1-x..0.1-x...2x
So you know I2 at equilibrium is 0.02M which means 0.1-x = 0.02. At equilibrium H2 must be the same. Calculate 2x = (HI).
Plug those values into kc expression and solve for Kc.
To calculate the value of Kc for a reaction, you will need the balanced chemical equation for the reaction and the initial and equilibrium concentrations of the reactants and products.
The balanced chemical equation for the reaction between hydrogen gas (H2) and iodine gas (I2) to form hydrogen iodide (HI) is:
H2(g) + I2(g) ⇌ 2HI(g)
Given:
Initial concentration of H2: 0.1 moles per liter
Initial concentration of I2: 0.1 moles per liter
Equilibrium concentration of I2: 0.02 moles per liter
The equilibrium concentrations of H2 and HI can be determined by using the stoichiometry of the balanced equation. Since the stoichiometric coefficient of I2 is 1, the equilibrium concentration of HI will also be 0.02 moles per liter. Using these values, we can now calculate the equilibrium concentration of H2.
Step 1: Set up an ICE table (Initial, Change, Equilibrium)
H2(g) + I2(g) ⇌ 2HI(g)
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I 0.1 -x 0.02
I2 0.1 -x 0.02
HI 0 +x 2x
Step 2: Calculate the change in concentration (x) for iodine (I2).
Initial concentration of I2 = 0.1 moles per liter
Equilibrium concentration of I2 = 0.02 moles per liter
Change in I2 concentration = Initial concentration - Equilibrium concentration
= 0.1 - 0.02
= 0.08 moles per liter
Step 3: Calculate the equilibrium concentration of H2.
Since the stoichiometric coefficient of H2 is 1, the change in H2 concentration will also be equal to x.
Change in H2 concentration = x
Equilibrium concentration of H2 = Initial concentration - Change in H2 concentration
= 0.1 - x
Step 4: Substitute the values into the expression for Kc.
Kc = ([HI]^2) / ([H2] * [I2])
Kc = (2x^2) / ((0.1 - x) * (0.08))
Step 5: Substitute values for x and solve for Kc.
Substitute x = 0.08 into the expression for Kc:
Kc = (2 * 0.08^2) / ((0.1 - 0.08) * (0.08))
= 0.032 / (0.02 * 0.08)
= 0.03 / 0.0016
= 18.75
Therefore, the value of Kc for the reaction is 18.75.
To calculate the value of Kc for a reaction, we need to use the equilibrium concentrations of the reactants and products. In this case, the reaction is between hydrogen gas (H2) and iodine gas (I2), and it is given that the initial concentrations of both reactants are 0.1 moles per litre.
The balanced chemical equation for this reaction is:
H2(g) + I2(g) ⇌ 2HI(g)
At equilibrium, let's assume that x moles of hydrogen gas and iodine gas have reacted to form 2x moles of hydrogen iodide gas. Therefore, the equilibrium concentrations will be:
[I2] = 0.1 - x moles per litre
[H2] = 0.1 - x moles per litre
[HI] = 2x moles per litre
Given that the concentration of iodine at equilibrium is 0.02 moles per litre, we can set up the following equation:
0.02 = 0.1 - x
Solving this equation, we find that x = 0.08 moles per litre. Substituting this value back into the equilibrium concentrations:
[I2] = 0.1 - 0.08 = 0.02 moles per litre
[H2] = 0.1 - 0.08 = 0.02 moles per litre
[HI] = 2 * 0.08 = 0.16 moles per litre
Now, we can calculate the value of Kc using the equilibrium concentrations:
Kc = ([HI]^2) / ([H2] * [I2])
= (0.16^2) / (0.02 * 0.02)
= 4 / 0.0004
= 10,000
Therefore, the value of Kc for this reaction is 10,000.