10.0 moles of NH3 and 10.0 moles of H2 are introduced into a 1.00 L flask. At equilibrium, 3.0 moles of NH3 gas remain. Calculate Kc for this reaction.
N2(g) + 3H2(g) < > 2NH3(g)
sorry i forgot to say show all the work please!
We usually don't do the whole problem. The idea on this board is to help you do your own homework, not to do it for you. Here are some hints.
You have the equation. Write the Kc expression. From the knowledge that 3.0 mols NH3 gas remains at equilibrium, that must mean that 7 mol NH3 must have been used. Use that to determine equilibrium concentrations for N2 and H2, and plug into Kc equation. If this hint is not enough, please post your work and tell us exactly what you don't understand about it.
Five (5.00) grams of glucose, C6H12)6, is dissolved in 500.0 grams of acetic acid. What is the new freezing point and boiling point for the solution?
Kf acetic acid = 3.90, Kb acetic acid = 3.07
(normal freezing point for acetic acid = 16.60 dg C, boiling point = 118.5 dg C)
f.p. = ??????
b.p. = ??????
To determine the new freezing point and boiling point of the solution, we need to use the equations for freezing point depression and boiling point elevation.
Freezing Point Depression:
ΔTf = Kf * molality
Boiling Point Elevation:
ΔTb = Kb * molality
First, we need to calculate the molality of the solution:
Moles of glucose (C6H12O6) = mass / molar mass
Molar mass of C6H12O6 = (6 * 12.01) + (12 * 1.008) + (6 * 16.00) = 180.18 g/mol
Moles of glucose = 5.00 g / 180.18 g/mol = 0.0278 mol
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of acetic acid solvent = 500.0 g = 0.500 kg
Molality = 0.0278 mol / 0.500 kg = 0.0556 mol/kg
Now, we can calculate the freezing point depression and boiling point elevation:
ΔTf = Kf * molality = 3.90 * 0.0556 = 0.216 dg C (rounded to 3 decimal places)
ΔTb = Kb * molality = 3.07 * 0.0556 = 0.170 dg C (rounded to 3 decimal places)
To find the new freezing point, subtract the freezing point depression from the normal freezing point of acetic acid:
f.p. = 16.60 dg C - 0.216 dg C = 16.38 dg C
To find the new boiling point, add the boiling point elevation to the normal boiling point of acetic acid:
b.p. = 118.5 dg C + 0.170 dg C = 118.67 dg C
Therefore, the new freezing point of the solution is 16.38 dg C and the new boiling point is 118.67 dg C.
To find the new freezing point and boiling point of the solution, we need to use the concept of colligative properties.
For the freezing point depression, we can use the equation:
ΔTf = Kf * m
Where:
ΔTf is the change in freezing point
Kf is the freezing point depression constant
m is the molality of the solution (moles of solute per kilogram of solvent)
First, we need to convert the given mass of glucose to moles. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol.
Moles of glucose = mass / molar mass
= 5.00 g / 180.16 g/mol
≈ 0.0278 mol
Next, we need to calculate the molality of the solution, which is the moles of solute per kilogram of solvent. The total mass of the solution is the sum of the mass of glucose and the mass of acetic acid.
Mass of solution = mass of glucose + mass of acetic acid
= 5.00 g + 500.0 g
= 505.0 g
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.0278 mol / (505.0 g / 1000)
≈ 0.055 mol/kg
Now we can calculate the freezing point depression (ΔTf) by plugging in the values:
ΔTf = Kf * m
= 3.90 * 0.055
≈ 0.2145 °C
To find the new freezing point, we subtract the freezing point depression from the normal freezing point of the solvent:
f.p. = normal freezing point - ΔTf
= 16.60 °C - 0.2145 °C
≈ 16.3855 °C
Therefore, the new freezing point of the solution is approximately 16.3855 °C.
For the boiling point elevation, we can use the equation:
ΔTb = Kb * m
Where:
ΔTb is the change in boiling point
Kb is the boiling point elevation constant
m is the molality of the solution (moles of solute per kilogram of solvent)
Using the same molality value (0.055 mol/kg) that we calculated earlier, we can calculate the boiling point elevation (ΔTb):
ΔTb = Kb * m
= 3.07 * 0.055
≈ 0.16835 °C
To find the new boiling point, we add the boiling point elevation to the normal boiling point of the solvent:
b.p. = normal boiling point + ΔTb
= 118.5 °C + 0.16835 °C
≈ 118.66835 °C
Therefore, the new boiling point of the solution is approximately 118.66835 °C.