onsider the following reaction: NO (g) + SO3 (g) <--->NO2 (g) + SO2 (g) A reacti

Question

onsider the following reaction:
NO (g) + SO3 (g) <--->NO2 (g) + SO2 (g)
A reaction mixture initially contains 0.86 atm NO and 0.86 atm SO3. Determine the
equilibrium pressure of NO2 if Kp for the reaction at this temperature is 0.0118

SO I KNOW .084 ATM IS THE ANSWER BUT, after getting =.10863 wouldn't we be left with .10863(.86-x)=x then we would multiply those values and move the X, wouldnt we be left with .0934=2x? and then dividing by 2x why is the answer not .047? what happened with the X and why didnt we get 2X on the other side?

NO (g) SO3 (g) NO2 (g) SO2 (g) I 0.86 0.86 +a +a -a E 0.86-a 0.86-a Kp P (NO)P(SO2)/P(NO)P(SO3) E a (0.86 a) J 0.01 18 a (0.86 a) 1/2 0.10863 0.0118 1.10863a 0.09342 a 0.0843 P (NO2) a 0.0843 atm z 0.084 atm

Explanation / Answer

a^2/(0.86-a)^2= 0.0118

a/(0.86-a) = 0.0118

a/(0.86 - a) = 0.10863

a=( 0.86-a)×0.10863

a= 0.09342 - 0.10863a

1.10863a = 0.09342

a = 0.09342/1.10863

= 0.084 atm

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