The reaction is 2 NO (g) <----> N2(g) + O2
partial pressures
Initial 37.30 0 0
Change -2p +p +p
Equilibrium 37.30-p p p
Kp = pN2 X pO2 / (pNO)^2
2400 = p^2 / (37.30-p)^2
3339096 - 179040p + 2400p^2 = p^2
2399p^2 + 3339096 -179040 p = 0
On solving
p = 36.55atm
Thus partial pressure of N2 and O2 = 36. 55 atm
Answer:
Ok:
Explanation:
So, you can use the Henderson-Hasselbalch equation for this:
pH = pKa + log(
) where A- is the conjugate base of the acid. In other words, A- is the deprotonated form and HA is the protonated.
We can solve that
1 = log(
) and so 10 = or 10HA = A-. For every 1 protonated form of adenosine (HA), there are 10 A-. So, the percent in the protonated form will be 1(1+10) or 1/11 which is close to 9 percent.
Reactants left products right
