At 25 degree C, the decomposition of dinitrogen tetraoxide N_2O_4(g) rightarrow

Question

At 25 degree C, the decomposition of dinitrogen tetraoxide N_2O_4(g) rightarrow 2NO_2(g) has an equilibrium constant (K_p) of 0.144. If the equilibrium pressure of nitrogen dioxide is 0.298 atm, what is the pressure of dinitrogen tetraoxide 2.07 atm 1.62 atm 0.617 atm 0.0128 atm Consider the following reactions at 700 degree C: 2SO_2(g) + O_2(g) rightarrow 2SO_3(g) K_1 = 4.8 2NO(g) + O_2(g) rightarrow 2NO_2(g) K_2 = 16 What is the equilibrium constant for the following reaction SO_2(g) + NO_2(g) rightarrow SO_3(g) + NO(g) K_3 = 21 1.8 11 77 0.55

Explanation / Answer

Solution :-

N2O4 ----- > 2NO2

Kp = 0.144

Equilibrium pressure of the NO2 = 0.298 atm

Equilibrium pressure of N2O4 = ?

Kp = [NO2]^2/[N2O4]

0.144 = [0.298]^2 / [x]

[x]= [0.298]^2 / 0.144

[x] = 0.617 atm

So the equilibrium pressure of the N2O4 = 0.617 atm so the answer is option d .

Q5).

The final equation is the sum of given two equations

To get the final equation we need to reverse the equation 2 and half both equations

When we reverse the equations then rate constant is inversed and when we half the equation then rate constant is taken as square root of the original rate constant

So for the final equation

K3 = (4.8)^0.5 * (1/16)^0.5

K3 = 0.55

So the answer is option e

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