Question

n which direction will the net reaction proceed.
X(g) + Y(g) <==> Z(g) .. Kp = 1.00 at 300k
for each of these sets of initial conditions?


1) [X] = [Y] = [Z] = 1.0 M
a] net reaction goes to the left [this one?]
b] net reaction goes to the right
c] reaction is at equilibrium

2. Px = Pz = 1.0 atm, Py = 0.50 atm

a] net reaction goes to the left
b] net reaction goes to the right
c] reaction is at equilibrium

Answer 1

Answer:

1) The correct answer is option b.

2) The correct answer is option a.

Explanation:

1)

$X+Y\rightleftharpoons Z$

At 300 K, the value of the $K_p=1.00$

The $K_c$ and $K_p$ is related by :

$K_p=K_c\times (RT)^{\Delta n}$

where,

$K_p$ = equilibrium constant at constant pressure

$K_c$ = equilibrium concentration constant

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 300 K

$\Delta n$ = change in the number of moles of gas = 1 - 2 = -1

Now put all the given values in the above relation, we get:

$1.00=K_c\times (0.0821 L atm/(K mol)\times 300)^{-1}$

$K_c=24.63$

The $K_c$ of the reaction = 24.63

Given = [X] = [Y] = [Z] = 1.0 M

Value of reaction quotient = Q

$Q=\frac{[Z]}{[X][Y]}=\frac{1.0 M}{1.0M\times 1.0 M}=1$

$Q<K_c$, the equilibrium will move in forward direction that is in the right direction.

2)

$X+Y\rightleftharpoons Z$

At 300 K, the value of the $K_p=1.00$

Given = $P_x = P_z = 1.0 atm, P_y = 0.50 atm$

Value of reaction quotient in terms of partial pressure = $Q_p$

$Q_p=\frac{P_z}{P_x\times P_y}=\frac{1.0 atm}{1.0 atm\times 0.50 atm}$

$Q_p=2$

$Q_p>K_p$the equilibrium will move in backword direction that is in the left direction.