When Compounds do not React in a 1:1 Molar Ratio The molar ratios affect how the concentrations change during a reaction. We must account for these ratios when we solve ICE box problems. Specifically they will affect the CHANGE line when we are doing these problems.

Practice Problem Nitrosyl bromide (NOBr) decomposes according to the reaction below: 2 NOBr  2 NO + Br M of NOBr is used initially and when the reaction reaches equilibrium, M NOBr remain. Determine the equilibrium constant! (K eq )

Start with the Initial Line! We know the starting concentration of NOBr. What about the concentrations of our products? Do we know any other concentrations? NOBrNOBr 2 Initial Change Equilibrium 0.410

Next is the Change Line! 2 NOBr  2 NO + 1 Br 2 We don’t know the exact change amount yet, but we know how they are going to change. Pay attention to the coefficients! NOBrNOBr 2 Initial Change - 2 X2 XX Equilibrium 0.410

Using the info for NOBr, we can solve for X, the change in concentration: M – 2x = M X = M But remember this is just X. We are looking for the equilibrium constant (K) so we need the rest of the values at equilibrium.. Solving for X NOBr Initial Change - 2 X Equilibrium 0.410

Solving the Concentrations at Equilibrium X = M (Previous Slide) Note: These numbers are plugged into the equation NOBrNOBr 2 Initial M00 Change Equilibrium M M0.090 M0.045 M M0.045 M - 2 X2 XX

2 NOBr  2 NO + Br 2 Remember coefficients when writing the equilibrium expression. K = [NO] 2 [Br 2 ] [NOBr] 2 K = [0.090] 2 [0.045] [0.410] 2 K = or 2.1x10 -3 NOBrNOBr 2 Equilibrium Plug in the values at equilibrium into the expression and solve for K:

Practice Problem – A different kind… Hydrogen gas reacts with Iodine gas to produce hydroiodic acid. Write and balance the chemical equation: 1 H I 2  2 HI Calculate all concentrations at equilibrium when initially M of both reactants are used and the equilibrium constant is equal to What makes this problem different than previous ones? – We are given the K eq value and asked to solve for the concentrations.

Start with the Initial! Again, this one is different because we are not given any concentrations at equilibrium. We are going to solve for that upcoming. H2H2 l2l2 Hl Initial Change Equilibrium ??

Next Comes the Change! 1 H I 2  2 HI Make sure you’ve balanced the equation first! H2H2 l2l2 Hl Initial Change -X-X-X-X2 X Equilibrium

Finally the Equilibrium! To fill in the rest of the boxes, it is always: INITIAL + CHANGE = EQUILIBRIUM Great! Now it’s time for us to solve for X ! H2H2 l2l2 Hl Initial Change -X-X-X-X2 X Equilibrium – X 2 X

Wait a second… There is no column for us to solve for X! Fear not my downhearted disciples of chemistry! We were given the K eq intially in the problem. We will write equilibrium expression and use the Keq to solve for X. Then plug that in to get the concentrations at equilibrium. H2H2 l2l2 Hl Equilibrium – X 2 X

1 H I 2  2 HI Remember coefficients when writing the equilibrium expression. K = [HI] 2 [H 2 ][I 2 ] 64.0 = [2X] 2 [0.200 – X][0.200 – X] H2H2 l2l2 Hl Equilibrium0.200 – X 2 X Plug in the values at equilibrium, and K eq = 64.0 into the expression and solve for X:

Let’s clean this up a little bit… In the bottom, two identical things are multiplied together. That means we can ??? So what can we do to get X by itself? Take the square root! 8.0 = 2X – X 64.0 = [2X] 2 [0.200 – X] 2  

Let’s clean this up a little bit… Need X, so multiply both sides by??? 8.0 = 2 X – X 8.0 (0.200 – X) = 2 X *Distribute Through* 1.6 – 8 X = 2 X *Combine Like Terms* 1.6 = 10 X *Solve for X!* 0.16 = X But remember that’s not the final answer. That’s just X, now lets plug it into the expression.

Yes, That is My Final Answer! H2H2 l2l2 Hl – X 2 X From the last slide, we solved for X = Lets Plug it in! And that’s it! At equilibrium there will be [0.040] H 2, [0.040] I 2, [0.32] HI – – (0.16)0.32