Question

For the reaction N2O4(g) m 2 NO2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (see the defi nition of standard state in Section 6.9 ) . If Kp = 0.15, which statement is true of the reaction mixture before any reaction occurs? (a) Q = K; the reaction is at equilibrium. (b) Q 6 K; the reaction will proceed to the right. (c) Q 7 K; the reaction will proceed to the left.

Answer 1

Answer:

c) Q > K, the reaction will proceed to the right

Explanation:

The given reaction is:

$N2O4(g)\rightleftharpoons 2NO2(g)$

The direction in which the above reaction will shift is based on the relation between the values of the equilibrium constant (K) and the reaction quotient (Q)

If Q = K, the reaction is at equilibrium

If Q < K, the reaction shifts to the left

If Q > K, the reaction shifts to the right

For a gas phase reaction, Q is given as:

$Qp = \frac{Products}{Reactants}$

Here,

$Qp = \frac{[NO2]^{2} }{[N2O4]} \\\\Under\ standard\ state\ : \\\\Pressure\ NO2 = \ Pressure\ N2O4\ = 1\ atm\\\\Qp = \frac{[1.00]^{2} }{[1.00]} = 1$

Kp = 0.15

Qp = 1.00

Since Qp > Kp, the reaction will shift to the right