1) Consider the following reaction where K c = 7.00×10 -5 at 673 K. NH 4 I (s) -

Question

1) Consider the following reaction where Kc = 7.00×10-5 at 673 K.
NH4I(s) --> NH3(g) + HI(g)
A reaction mixture was found to contain 5.62×10-2 moles of NH4I(s), 1.12×10-2 moles of NH3(g), and 8.37×10-3 moles of HI(g), in a 1.00 liter container.

Is the reaction at equilibrium?
If not, what direction must it run in order to reach equilibrium?
The reaction quotient, Qc, equals (???????)
The reaction ?????
A. must run in the forward direction to reach equilibrium.
B. must run in the reverse direction to reach equilibrium.
C. is at equilibrium.

2) The equilibrium constant, Kp, for the following reaction is 8.46×10-5 at 298 K:
NH4HSe(s) --> NH3(g) + H2Se(g)
Calculate the partial pressure of each gas and the total pressure at equilibrium when 0.449 moles of NH4HSe(s) is introduced into a 1.00 L vessel at 298 K. Assume that the volume occupied by the solid is negligible.

PNH3 =  atm
PH2Se =  atm
Ptotal =  atm

3) The equilibrium constant, Kp, for the following reaction is 0.444 at 452 K:
(CH3)2CHOH(g) --> (CH3)2CO(g) + H2(g)
Calculate the equilibrium partial pressures of all species when (CH3)2CHOH(g) is introduced into an evacuated flask at a pressure of 1.35 atm at 452 K.

Explanation / Answer

Ans 1

The given reaction

NH4I(s) --> NH3(g) + HI(g)

Equilibrium constant

Kc = 7.00×10^-5 at 673 K

[NH4I] = 5.62*10^-2 M

[NH3] = 1.12*10^-2 M

[HI] = 8.37*10^-3 M

Reaction Quotient

Q = [HI] [NH3] / [NH4I]

= (8.37*10^-3) * (1.12*10^-2) / (5.62*10^-2)

= 1.67*10^-3

Kc (7.00×10^-5) < Q (1.67*10^-3)

The reaction is not at equilibrium

The reaction proceeds in backward direction

B. must run in the reverse direction to reach equilibrium

Option B is the correct answer

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