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3 years ago

Convert 180 g/mol to mole/L

Chemistry

1 answer:

jarptica [38.1K]3 years ago

7 0

180, 000, 000, m-3 kg-1 mol2

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3 years ago

What is the numerical value of the equilibrium constant Kc?

Nezavi [6.7K]

ANSWER

EXPLANATION

Given that;

The number of moles of NH3 is 3 moles

The number of moles of H2 is 1 mole

The number of moles of N2 is 2 moles

At equilibrium, the concentration of ammonia is 1.4 moles/L

To find the value of Kc, follow the steps below

Step 1: Write the balanced equation of the reaction

$\text{ N}_{2(g)}\text{ + 3H}_{2(g)}\text{ }\rightleftarrows\text{ 2NH}_{3(g)}$

Step 2: Write the equation of the reaction in terms of Kc

$\text{ K}_C\text{ = }\frac{[\text{ NH}_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$

Step 3: Find the concentration of each reactant at equilibrium using a stoichiometry ratio

From the reaction above, you will see that 1 mole of Nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.

Let x represents the concentration of nitrogen at equilibrium

Recall, that the concentration of ammonia at equilibrium is 1.4 moles/L

$\begin{gathered} \text{ 2 mole NH}_3\text{ }\rightarrow\text{ 1 mole N}_2 \\ \text{ 1.04 mole/L NH}_3\text{ }\rightarrow\text{ x moles/L N}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ x moles/L N}_2\text{ = 1 mole N}_2\times1.4\text{ mol/L} \\ \text{ Isolate x} \\ \text{ x mol/L N}_2\text{ = }\frac{1\text{ moles N}_2\times1.41\cancel{\frac{mol}{L}}}{2\cancel{moles}} \\ \text{ x mol/L = }\frac{1.04}{2} \\ \text{ x= 0.52 mole/L} \end{gathered}$

Since the initial number of moles of nitrogen is 1 mole, hence, the concentration of nitrogen at equilibrium is calculated below as

Concentration at equilibrium = 1 -0.52

Concentration of nitrogen at equilibrium = 0.48 mole/L

The next step is to find the concentration of hydrogen at equilibrium

Let y represent the mole of hydrogen at equilibrium

$\begin{gathered} \text{ 2 moles NH}_3\rightarrow\text{ 3 moles H}_2 \\ \text{ 1.04 moles/L NH}_3\text{ }\rightarrow\text{ y moles/L H}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ y moles/L H}_2\text{ = 1.04 moles/L NH}_3\times3\text{ moles H}_2 \\ \text{ Isolate y} \\ \text{ y moles/L H}_2\text{ = }\frac{1.04\cancel{\frac{moles}{L}}NH_3\times3mole\text{ H}_2}{2\cancel{molsNH_3}} \\ y\text{ = }\frac{1.40\times3}{2} \\ \text{ y = }\frac{3.12}{2} \\ \text{ y = 1.56 moles} \end{gathered}$

Since the initial concentration of hydrogen is 2 moles, hence the concentration of hydrogen at equilibrium can be calculated below as

Concentration at equilibrium = 2 - 1.56

Concentration at equilibrium = 0.44 mole/L

Step 4: Find the value of Kc using the equation in step 2

$\text{ Kc = }\frac{[NH_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$ $\begin{gathered} \text{ Kc = }\frac{1.04^2}{0.48\times0.44^3} \\ \\ \text{ K}_c=\text{ }\frac{1.0816}{0.48\times\text{ 0.085184}} \\ \\ \text{ kc = }\frac{1.0816}{0.0408832} \\ \text{ kc = 26.4558547276} \end{gathered}$

Hence, the value of Kc is 26

6 0

2 years ago

2. How will the equilibrium shift if the following changes are made? State if the reaction will shift

Kamila [148]

Please give me brainleist. :)

Answer:

2a. If the temperature is increased, the reaction will shift to the right in an attempt to release some of the heat. As the forward reaction loses heat while the reverse would create more heat.

2b. If the pressure is increased, it would shift to the left to counteract the increase in pressure as the left side will have fewer molecules.

2c. If Cl2 is added the reaction will shift to the left in order to remove the stress of the extra Cl2 and favor the production of more reactant.

2d. If PCl3 is removed, the reaction will shift to the right. When part of the equation is removed the reaction learns to adapt to the loss by trying to make more Pcl3 and counteract the effects of losing the PCl3.

3a. The reaction will shift to the right to produce more heat and counter the negative effects of losing the heat.

3b. It will shift to the left to get rid of the excess HCl being produced and form more reactant from the breakdown of the HCl.

3c. It would shift to the right in order to get rid of the excess form products from it.

3d. If pressure is decreased there will be no effect on the shift of the reaction because there is an even amount of moles of gas on each side.

4a. K=[N2O4(g0] / [NO2(g)]2

4b. (Below)

K=[N2O4(g)] / [NO2(g)]2

0.4 / 0.5(2)

0.4/0.25 = 1.6

Keq= 1.6

7 0

3 years ago