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Nady [450]
3 years ago
13

Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are

added. at equilibrium, 0.100 mol of nh3 is present. calculate the equilibrium concentrations of n2 and h2 for the reaction: 2 nh3(g) n2(g) + 3 h2(g). n2 1.25 mol/l h2 1.50 mol/l find kc for the reaction. kc = 422 m2 (b) in a different 1.00-l container at the same temperature, equilibrium is established with 8.34 ✕ 10−2 mol of nh3, 1.50 mol of n2, and 1.25 mol of h2 present. calculate kc for the reaction: nh3(g) 1 2 n2(g) + 3 2 h2(g). 20.5 m (c) what is the relationship between the kc values in parts (a) and (b)? why aren't these values the same?
Chemistry
1 answer:
emmasim [6.3K]3 years ago
8 0
When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

      2 NH₃ ⇄ N₂ + 3 H₂
I         0        1.3    1.65
C     +2x       -x      -3x
-------------------------------------
E       0.1        ?        ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
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<u>Answer:</u> The molality of the solution is 0.11 m

<u>Explanation:</u>

We are given:

Mole fraction of methanol = 0.135

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Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

Where,

m_{solute} = Given mass of solute (methanol) = 0.135 g

M_{solute} = Molar mass of solute (methanol) = 32 g/mol

W_{solvent} = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m

Hence, the molality of the solution is 0.11 m

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