13.5ml of hydrogen iodide are produced by the reaction of 8.1ml of hydrogen and 9.3ml of iodine vapor at444°c. Calculate the Kc at this temperature
Well, chemistry can be quite "heated," can't it? Let's calculate that Kc for you!
To find the Kc, we need to use the balanced equation for the reaction and determine the concentrations or partial pressures of the reactants and products.
The balanced equation for the reaction is:
H2(g) + I2(g) ⇄ 2 HI(g)
Now, let's calculate the concentrations of H2, I2, and HI.
First, let's convert the volumes to moles using the ideal gas law (PV = nRT). However, remember that the volume of the reactants and products should be expressed at the same temperature and pressure. Since it's not given in the question, let's assume they are at standard temperature and pressure conditions.
Using the molar volume at standard temperature and pressure (22.4 L/mol), we get:
H2: 8.1 mL -> 0.0081 L -> (0.0081/22.4) mol
I2: 9.3 mL -> 0.0093 L -> (0.0093/22.4) mol
HI: 13.5 mL -> 0.0135 L -> (0.0135/22.4) mol
Now, let's plug these values into the equation:
Kc = [HI]^2 / ([H2] * [I2])
Kc = [(0.0135/22.4)^2] / [(0.0081/22.4) * (0.0093/22.4)]
Now, let's calculate that bad boy!
Kc = [(0.000604)^2] / [(0.000363) * (0.000415)]
Kc ≈ 6.61
So, the Kc at this temperature is approximately 6.61.
If that didn't make you smile, I hope the chemistry behind it at least cleared the air!
To calculate the equilibrium constant (Kc) at this temperature, we need to determine the initial concentrations of the reactants and products.
Given:
- Volume of hydrogen (H2) = 8.1 ml
- Volume of iodine (I2) vapor = 9.3 ml
- Volume of hydrogen iodide (HI) produced = 13.5 ml
First, we need to convert the volumes to concentrations. Assuming that the volume ratios are equivalent to the molar ratios since the reaction follows the stoichiometry:
- Concentration of H2 (initial) = 8.1 ml / 22.4 ml/mol (molar volume at STP) = 0.362 moles/liter
- Concentration of I2 (initial) = 9.3 ml / 22.4 ml/mol (molar volume at STP) = 0.415 moles/liter
- Concentration of HI (initial) = 13.5 ml / 22.4 ml/mol (molar volume at STP) = 0.603 moles/liter
Now, we can express the balanced equation for the reaction as follows:
H2 + I2 ⇌ 2HI
Using the stoichiometry, we can write the equilibrium concentrations as follows:
[H2] = (initial concentration of H2) - x
[I2] = (initial concentration of I2) - x
[HI] = (initial concentration of HI) + 2x
Where x represents the change in concentration at equilibrium.
Since the reaction produces 13.5 ml of HI, and each mole of HI has a volume of 22.4 ml, we can determine the value of x:
(initial concentration of HI) + 2x = 13.5 ml / 22.4 ml/mol
0.603 + 2x = 0.603 moles/liter
2x = 0.603 - 0.603 = 0
x = 0
Therefore, the equilibrium concentrations are:
[H2] = 0.362 moles/liter
[I2] = 0.415 moles/liter
[HI] = 0.603 moles/liter
The equilibrium expression for Kc is given by:
Kc = [HI]² / ([H2] * [I2])
Plugging in the values:
Kc = (0.603)² / (0.362 * 0.415)
Kc = 0.3641 / 0.15013
Kc ≈ 2.426
Therefore, the equilibrium constant (Kc) at this temperature is approximately 2.426.
To calculate the equilibrium constant (Kc) for this reaction, we first need to determine the molar concentrations of the reactants and products at equilibrium.
Step 1: Calculate the moles of hydrogen (H2) and iodine (I2) using the given volumes and the ideal gas law.
Using the ideal gas law equation, PV = nRT where:
P = pressure (assuming constant and at equilibrium)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature
For hydrogen gas:
P = assuming constant (could be obtained from additional information)
V = 8.1 ml = 0.0081 L
T = 444°C + 273.15 = 717.15 K
Using PV = nRT, we can calculate the moles of hydrogen:
n(H2) = (P × V) / (R × T)
For iodine vapor:
P = assuming constant
V = 9.3 ml = 0.0093 L
T = 444°C + 273.15 = 717.15 K
Using PV = nRT, we can calculate the moles of iodine:
n(I2) = (P × V) / (R × T)
Step 2: Determine the initial and equilibrium concentrations.
Initial moles and volumes:
n(H2) = calculated in Step 1
V(H2) = 8.1 ml = 0.0081 L
n(I2) = calculated in Step 1
V(I2) = 9.3 ml = 0.0093 L
n(HI) = 0 (since no HI is given initially)
V(HI) = 0.0135 L
Molar concentrations at equilibrium:
[H2] = (n(H2) / V(H2))
[I2] = (n(I2) / V(I2))
[HI] = (n(HI) / V(HI))
Step 3: Write the balanced chemical equation for the reaction.
H2(g) + I2(g) ⇌ 2HI(g)
Step 4: Use the balanced chemical equation to determine the relationship between the concentrations at equilibrium.
Kc = ([HI]^2 / ([H2] × [I2]))
Step 5: Substitute the values obtained in Step 2 into the equation derived in Step 4.
Kc = ([HI]^2 / ([H2] × [I2]))
Kc = ([0.0135]^2 / ([n(H2) / V(H2)] × [n(I2) / V(I2)]))
Kc = ([0.0135]^2 / ([n(H2) / 0.0081] × [n(I2) / 0.0093]))
Step 6: Substitute the calculated moles from Step 1 into the equation derived in Step 5.
Kc = ([0.0135]^2 / (n(H2) / 0.0081) × (n(I2) / 0.0093))
After obtaining the values for n(H2) and n(I2) from Step 1, you can plug them into the equation to calculate Kc at the given temperature.
mols H2, use PV = nRT and calculate n.
mols I2 do the same and find n.
Then initial (H2) = n/L and initial (I2) = n/L
Do you have a pressure? or some other way to calculate n?
....H2 + I2 ==> 2HI