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.
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!!
3 answers
I am sorry it is in grams.
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.
(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.
1 answer