At 1285 degree C the equilibrium constant for the reactin Br2(g) <> 2Br(g) is Keq= 1.04 * 10^-3. A 0.200 L vessel containing an equilibrium mixture of the gases has 0.245 Br2(g) in it. What is the mass of Br(g) in the vessel?
Please help, Thanks!!
At 1285 degree C the equilibrium constant for the reactin Br2(g) <> 2Br(g) is Keq= 1.04 * 10^-3. A 0.200 L vessel containing an equilibrium mixture of the gases has 0.245 WHAT Br2(g) in it. What is the mass of Br(g) in the vessel?
No unit is listed. That could be moles, molarity, g/L, just what.
Br2(g) <> 2Br(g)
(Br2) = moles/L =
moles = g/molar mass = 0.245 g/159.81 = 0.00153 and the volume is 0.2 L; therefore,
(Br2) = 0.00153/0.200 = 0.00766 at equilibrium.
Substitute into Keq expression and solve for (Br) and that will be in moles/L. Multiply that by 0.200 L to convert to moles and by atomic mass Br to convert to grams (although g Br doesn't make sense to me since Br doesn't exist as monatomic gas (I guess it might at umpteen thousand degrees K0.
To find the mass of Br(g) in the vessel, we need to use the given equilibrium constant (Keq) and the initial amount of Br2(g) in the vessel.
First, let's convert the given volume of the vessel to moles of Br2(g). To do this, we need to know the molar volume of an ideal gas at the given conditions. Without this information, we will assume that the gas behaves ideally and use the ideal gas law to calculate the moles of Br2(g).
The ideal gas law is given by the formula:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
Since we are not given the pressure or temperature, we will also make assumptions to proceed with the calculations. Let's assume the pressure is 1 atm and convert the given temperature of 1285 degrees Celsius to Kelvin by adding 273.15.
Now, let's calculate the moles of Br2(g):
n = PV / RT
Substituting the values:
P = 1 atm
V = 0.200 L
R = ideal gas constant (0.0821 L.atm/mol.K)
T = 1285°C + 273.15 = 1558.15 K
n = (1 atm * 0.200 L) / (0.0821 L.atm/mol.K * 1558.15 K)
Now we have the number of moles of Br2(g) present in the vessel.
Next, we need to use the equilibrium constant (Keq) to calculate the moles of Br(g). According to the stoichiometry of the reaction, 2 moles of Br(g) is formed per mole of Br2(g) consumed. Therefore, the moles of Br(g) is twice the moles of Br2(g) consumed.
moles of Br(g) = 2 * moles of Br2(g)
Now that we have the moles of Br(g), we can convert it to mass using the molar mass of Br (79.90 g/mol).
mass of Br(g) = moles of Br(g) * molar mass of Br
Finally, we have calculated the mass of Br(g) in the vessel.
Please note that this approach assumes ideal gas behavior and makes assumptions about the pressure and temperature in the absence of specific information. Additionally, it assumes that the initial moles of Br2(g) are fully consumed and no other reactions or equilibrium shifts occur in the vessel.