If the equilibrium constant Kc for the reaction below...

If the equilibrium constant $K_c$ for the reaction below is 1.71 × 10$^{-1}$, what will be the equilibrium pressure of NO if the initial partial pressures of the three gases are all 1.86 × 10$^{-3}$ atm? $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$

Transcript

00:01 Determine the equilibrium pressure of the no nitrogen monoxide for this reaction and we are given this information.

00:12 We are given a kc but we are given pressures of gases so our starting pressures for all of these for all three gases is 0 .00176.

00:31 Okay so since we're given pressures for gases we can't use kc instead we need to use kp or find kp.

00:40 So kp equals kc times rt to the delta n.

00:44 R is our gas constant, t is our temperature.

00:47 We aren't given the temperature at all for this however delta n is our change in moles of gas from one side to the other.

00:56 So if we look at this reaction up here at the top we have one mole of n2, one mole of o2, and two moles of no.

01:04 So 2 minus 2 is 0 so in this case then delta n equals 0.

01:10 Anything to the 0 power equals 1 so that would mean that this also is the value for our kp.

01:18 Okay so that was the first battle.

01:20 So then the next battle is to figure out which way this reaction will shift since we have all three gases initially.

01:29 So we have to calculate q and we are going to compare k and q.

01:36 Let me fix this here.

01:37 We are going to calculate k and q.

01:41 So k is 0 .261.

01:46 Q is going to equal our products over our reactants raised to their powers.

01:53 So we would have concentration of no squared over the concentration of n2 to the first and o2 to the first.

02:03 Now that means it's going to be 0 .00176 squared over 0 .00176 squared which means q equals 1.

02:16 So then we can figure out which way this reaction will shift.

02:21 If you compare these if you always put them k before q alphabetical order like i have here then this arrow the greater than less than sign points the direction that gets the plus x's.

02:35 So our change for this is going to be plus x.

02:38 Our change for this will be plus x and our change for this will be minus 2x.

02:45 So if we write this in terms of plus x that's in terms of x 0 .00176 plus x and then here 0 .00176 whoops minus 2x.

03:04 Then we can set this up...

Question

If the equilibrium constant $K_c$ for the reaction below is 1.71 × 10$^{-1}$, what will be the equilibrium pressure of NO if the initial partial pressures of the three gases are all 1.86 × 10$^{-3}$ atm? $N_2(g) + O_2(g) \rightleftharpoons 2NO(g)$

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