If we had 6 moles of C2H6(g) and 6 moles of Cl2(g) originally in a 3 liter container of 10degree Celsius, determine the equilibrium concentration. K for this reaction at 10degree Celsius is 0.10
the equation goes like this (it's not yet balanced though): C2H6(g) + Cl2(g) ---> C2H2Cl + HCl
i tried balancing it( but i don't know if its correct) : 2C2H6(g) + 5Cl2(g) ---> 2C2H2Cl + 8HCl
the equation goes like this (it's not yet balanced though): C2H6(g) + Cl2(g) ---> C2H2Cl + HCl
i tried balancing it( but i don't know if its correct) : 2C2H6(g) + 5Cl2(g) ---> 2C2H2Cl + 8HCl
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2C2H6(g) + 5Cl2(g) ---> 2C2H2Cl + 8HCl is balanced.
We're going to use what's known as an ICE table (which you'll probably learn about very soon(and I'm not going to explain because they require lines, and Answers is a terrible place for them))
Your initial concentration of C2H6 is 2M (6moles/L (M is molarity if you did't know that, and is expressed in moles/liter)) your initial concentration of Cl2 is also 2M, you have no C2H2Cl or HCl, so both of their concentrations a 0M.
We know that the concentrations are going to change during this reaction, so we're going to assign the change 1 mole of a substance would undergo as "x". Because we defined that as the change of one mole, we have to multiply the change based on the number of moles of each substance (which is given by the equation, not your initial values).
So the final concentration (the initial+the change) is 6-2x for C2H6 and 6-5x for Cl2 (these are subtractions because these two are reacting to form the product) the concentration for C2H2Cl and HCl are 2x and 8x respectively.
So now to the real math.
K=[Products]/[reactants] (the brackets indicate concentration)
K=[C2H2Cl]^2[HCl]^8/[C2H6]^2[Cl2]^5 Now, because this is a closed container and volume is constant, the volume terms for all of the elements cancel, and we can just deal with the number of moles.
So,
.10=(2x)^2(8x)^8/(6-2x)^2(6-5x)^2
so x=.251moles.
therefore, C2H2Cl=.502moles HCl=2.008moles Cl2=4.745moles C2H6=5.498moles
We're going to use what's known as an ICE table (which you'll probably learn about very soon(and I'm not going to explain because they require lines, and Answers is a terrible place for them))
Your initial concentration of C2H6 is 2M (6moles/L (M is molarity if you did't know that, and is expressed in moles/liter)) your initial concentration of Cl2 is also 2M, you have no C2H2Cl or HCl, so both of their concentrations a 0M.
We know that the concentrations are going to change during this reaction, so we're going to assign the change 1 mole of a substance would undergo as "x". Because we defined that as the change of one mole, we have to multiply the change based on the number of moles of each substance (which is given by the equation, not your initial values).
So the final concentration (the initial+the change) is 6-2x for C2H6 and 6-5x for Cl2 (these are subtractions because these two are reacting to form the product) the concentration for C2H2Cl and HCl are 2x and 8x respectively.
So now to the real math.
K=[Products]/[reactants] (the brackets indicate concentration)
K=[C2H2Cl]^2[HCl]^8/[C2H6]^2[Cl2]^5 Now, because this is a closed container and volume is constant, the volume terms for all of the elements cancel, and we can just deal with the number of moles.
So,
.10=(2x)^2(8x)^8/(6-2x)^2(6-5x)^2
so x=.251moles.
therefore, C2H2Cl=.502moles HCl=2.008moles Cl2=4.745moles C2H6=5.498moles
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keywords: really,help,need,this,prof,us,teach,your,Our,039,didn,Hey,yet,don,guys,know,Hey guys!. I need your help. :D I really don't know. Our prof didn't teach us this yet.