A mixture of 0.10 mol of NO, 0.050 mol of H_2, and 0.10 mol of H_2O is placed in
ID: 487011 • Letter: A
Question
A mixture of 0.10 mol of NO, 0.050 mol of H_2, and 0.10 mol of H_2O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established: At equilibrium [NO] = 0.062 M. (a) Calculate the equilibrium concentrations of H_2. N_2, and H_2O. (b) Calculate K_c. For the reaction: at 100 degree C. At a point during the reaction, [N_2O_4] = 0.12 M and [NO_2] = 0.55 M. Is the reaction at equilibrium? If not in which direction is it progressing? Consider. How does each of the following changes affect the yield of NO at equilibrium? (Answer increase, decrease or no change) increase in [NH_3] increase in [H_2O] decrease in [O_2] decrease the volume of the container in which the reaction occurs add a catalyst increase temperature Gaseous hydrogen iodide is placed in a closed container at 425 degree C, where it partially decomposes to hydrogen and iodine: At equilibrium it is found that [HI] = 3.53 times 10^-3 M, [H_2] = 4.79 times 10^-4 M, and [I_2] = 4.79 times 10^-4 M. What is the value of K_c at this temperature?Explanation / Answer
5) the reaction is
2 NO + 2H2 -----------> N2 + 2H2O
0.1 0.05 0 0.1 initial concentrations
0.1-2x 0.05-2x x 0.1+2x equilibrium moles
Given {NO} at equilibrium = 0.062 = 0.1-2x
thus x = 0.019
0.062 0.012 0.019 0.138 equilibrium concentrations
Thus Kc = [0.019][0.138]2 / [0.062]2[0.012]2
= 653.68
Q6) To know whether a reaction is at equilibrium or not , we calculate the reaction quotient Qc
If Qc> Kc the rection goes reverse
If Qc < Kc the reaction goes in forward direction. If Qc = Kc it is at equilibrium
Qc for this reaction = [NO2]2/ [N2O4}
= [0.55]2 / 0.12
=2.52
the Kc of this reaction i s0.21
As Qc > Kc , the rection goes in reverse direction.
Q7) The reaction is exothermic and proceeds with more number o f moles of gaseous products than rectants.
thus
a)increase in NH3 - reactant concentration increases , equilibrium shifts to right [NO] increases.
b) Increase in [H2O] - product increases, the reaction goes backwards , thus [NO] decreases.
c) decrease in [O2] -- reactnat decreases, the equilibrium goes backward.thus [NO] decreases.
d) volume of the container decreased - overall pressure increases, the n the equilibrium shifts to the side where less number of gaseous moles present. here it goes backward. thus [NO] decreases
e) addition of catalyst - catalyses both forward and reverse reactions equally. No efect on equlibrium and [NO]
f) increase in temperature - As the reaction is exothermic, increase in temperature favors backward reactio, thus [NO] decreases.
Q8) 2HI <-------> H2 + I2
given [HI] = 3.53x10-3 [H2] = [I2] = 4.79x10-4 at equilibrium
Thus Kc = [H2][I2] /[HI]2
= 4.79x10-4 x4.79x10-4 / 3.53x10-3 x3.53x10-3
= 1.81x10-2
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