If the equilibrium constant kc 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 1.96 × 10–3 atm? $$

The equilibrium reaction that is not shown is the reaction of N₂ and O₂ to form NO:

N₂ + O₂ ⇄ 2NO

We can write an expression for the equilibrium constant. We are given the value of Kc, but we must use Kp if we are dealing with partial pressures. However, since 2 moles of gas are formed from 2 moles of gas, the value of Kc is equal to Kp so we do not need to change anything.

Kc = (Pno)²/(Pn2)(Po2) = 1.71 x 10⁻¹

We must first use the initial pressures to find the reaction quotient to decide which way the equilibrium will shift:

Q = (1.96 x 10⁻³)²/(1.96 x 10⁻³)(1.96 x 10⁻³) = 1

Since Q > Kp, the equilibrium will shift to the left. Now we can prepare an ICE table to find the equilibrium pressures:

           N₂                       O₂                    NO
I      1.96 x 10⁻³       1.96 x 10⁻³         1.96 x 10⁻³
C       +x                         +x                    -2x
E    0.00196 + x     0.00196 + x        0.00196 - 2x

We enter these equilibrium pressures into the equilibrium expression and solve for x:

0.171 = (0.00196 - 2x)²/(0.00196 + x)²

Expand the equation and simplify to a quadratic function:

3.829x² - 0.00851x + 3.183·10⁻⁶ = 0

x = ((0.00851) +/- sqrt((0.00851)² - 4(3.829)(3.183 x 10⁻⁶)))/2(3.829)

x = 0.000476 atm

Now we can solve for the equilibrium partial pressures:

Pno = 0.00196 - 2(0.000476) = 0.00101 atm

Pn2 = 0.00196 + 0.000476 = 0.00244 atm

Po2 = 0.000196 + 0.000476 = 0.00244 atm

Therefore, the equilibrium partial pressure of NO is 0.00101 atm.

The answer is neonhope it