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A reaction container contains 10.7 atm of both N2 and O2 and 5.97 atm of NO....

A reaction container contains 10.7 atm of both N2 and O2 and 5.97 atm of NO. What is the pressure of NO once equilibrium is established? Kp = 2.4 x 10-2

N2 (g) + O2(g) ⇌ 2 NO (g)

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Expert Solution

                    N2 (g) + O2(g) ⇌ 2 NO (g)

I                 10.7     10.7           5.97

C                 +x          +x                -2x

E               10.7+x   10.7+x          5.97-2x

   Kp    = PNO^2/PN2PO2

2.4*10^-2   = (5.97-2x)^2/(10.7+x) *(10.7+x)

2.4*10^-2   = [(5.97-2x)/(10.7+x)]^2

0.155        = (5.97-2x)/(10.7+x)

0.155*(10.7+x) = (5.97-2x)

x    = 2

PN2 = 10.7 +x = 10.7+2 = 12.7atm

PN2 = 10.7 +x = 10.7+2 = 12.7atm

[NO]   = 5.97-2x   = 5.97-2*2 = 1.97atm

queen_honey_blossom answered 2 years ago
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