1 mole nitrogen gas and 3 moles hydrogen gas form .05 moles NH3 @ equilibrium. What is Kc of the reaction?

In a sealed 1 L container. I have (NH3)^3/((N2)(H2)^3).
The answer says:
1 mole N2 and 3 moles H2, some amount x of N2 will react with 3x amount of H2 to form 2x amount of ammonia, due to stoichiometric coefficients.
The amount of ammonia at equilibrium is 0.05 M, so 2x 0.05, and x = 0.025.
*Thus [N2] at equilibrium is 1-0.025 = 0.975 M, and [H2] is 3 - 3 x 0.025 M.

Why is anything subtracted from the stoichiometric coefficients at " * "?
Thanks

Question
Answer 1

#K_c~~1.02*10^-4#

The easiest method to organize all the information they are throwing at you is to separate all the parts into an initial, change, and equilibrium box (or ICE box). You want to first write out your chemical equation and balance it out. (unbalanced) #N_2+H_2hArrNH_3# (balanced) #N_2+3H_2 hArr 2NH_3#

Since Kc uses the concentrations of the substances, you want to divide all the moles given by the volume to give you molarity. Then organize all your given information and the inferred information to the table. The initial concentrations are #1 M_(N_2)# , #3M_(H_2)#, and #0M_(NH_3)#. We are also given an equilibrium amount of moles of #NH_3#, which converted into molarity is #.05M_(NH_3)#

(The commas are for spacing in the box) ,,,,,,,,,,,,,,,,#N_2#,,,,,,,,,#H_2#,,,,,,,,,,,,,,#NH_3# Initial,,,,,,,1,,,,,,,,,,,3,,,,,,,,,,,,,,,,,,,,0 Change,-x,,,,,,,,-3x,,,,,,,,,,,,,,,,+2x Equil.,,,,,1-x,,,,,,,3-3x,,,,,,,,,,,,+0.05

The change is the amount of reactants used to produce the products. In our chemical formula, we can infer that for every mole of #N_2# we use, 3 moles of #H_2# will be used and 2 moles of #NH_3# will be produced. With this same logic, if we use x moles of #N_2#, then we will have to use three times as much #H_2# and we will produce two times the amount we used of #N_2# as product, #NH_3#.

Since the concentration of #NH_3# started at 0 and changed to 0.05 at equilibrium and we know the change was +2x, we can say #0.05=0+2x# or #0.05=2x#. Then after solving for x (#x=0.025#), we can substitute the x into the equilibrium concentrations we have in our ICE box and get all the equilibrium concentrations. ,,,,,,,,,,,,,,,,,,,,,,#N_2#,,,,,,,,,,,,,,,,,,,,#H_2#,,,,,,,,,,,,,,,,#NH_3# Equil.,,,,,1-(0.025),,,3-3(0.025),,,,,,,,,,,,+0.05

,,,,,,,,,,,,,,,,#N_2#,,,,,,,,,#H_2#,,,,,,,,,,,,,,#NH_3# Equil.,,,,,0.975,,,,,,2.925,,,,,,,,,,,,0.05

We can then write our rate quotient/equilibrium constant (#K_c#) with our calculated equilibrium concentrations. #K_c=([NH_3]^2)/([N_2]*([H_2]^3))~~1.02*10^-4#

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