Question

1.Consider the following reversible reaction. In a 3.00 liter container, the following amounts are found in equilibrium at 400 oC: 0.0420 mole N2, 0.516 mole H2 and 0.0357 mole NH3. Evaluate Kc.N2(g) + 3H2(g) 2NH3(g)

Answer 1

Answer:

1.987

Explanation:

Given,

In a 3.00 liter container, the following amounts are found in equilibrium at 400 oC: 0.0420 mole N2, 0.516 mole H2 and 0.0357 mole NH3.

To Find,

Convert moles and volume to concentration:

So,

Concentration = moles / volume -> M

NH3 concentration = 0.0357 moles / 3 L = 0.0119 M

H2 concentration = 0.516 moles / 3 L = 0.172 M

N2 concentration = 0.0420 moles / 3 L = 0.014 M

Then,

Kc=([NH3]²) / ([H2]³[N2])

   = ([0.0119]²) / ([0.172]³[0.014] )

   = 1.987

Hence, 1.987 is the correct answer.

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Answer 2

Answer:

The equilibrium coefficient of the given reaction is found out to be 1.987.

Explanation:

Given - a 3L container
            the temperature of 400 °C
            0.0420 moles N₂
            0.0357 moles NH₃

            0.516 moles H₂

To find - the concentration coefficient of the given reaction $N_2 + 3H_2 \rightarrow 2NH_3$

Solution -

We know that the concentration of a given substance is the ratio of the number of moles of the substance to the volume of the given substance.

$Concentration \; (M)= \frac{moles}{volume}$

$M_N_H_3 = \frac{0.0357}{3} = 0.0119 M\\M_H_2 = \frac{0.516}{3} = 0.172 M\\\\ M_N_2 = \frac{0.0420}{3} = 0.014 M$

Thus, we can calculate the equilibrium coefficient of the given reaction using the following formula

$K_c = \frac{(NH_3)^2}{(N_2) (H_2)^3}\\\\K_c = \frac{(0.119)^2}{(0.172)^3 (0.014)}$

$K_c = 1.987$

Thus, the equilibrium coefficient of the given reaction is 1.987.

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