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mamaluj [8]
3 years ago
10

If the reaction N2 (g) + 3 H2 (g) --> 2 NH3 (g) has the concentrations 1.1 M for nitrogen, 0.75 M for hydrogen and 0.25 M fo

r ammonia gas, what is the Kc? Show all work. Does this mean that there are more reactants or products at equilibrium? Explain how you determined that.
Chemistry
1 answer:
Luba_88 [7]3 years ago
4 0

<u>Answer:</u> The value of K_c is 0.136 and is reactant favored.

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{c}

For the chemical reaction between carbon monoxide and hydrogen follows the equation:

N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)

The expression for the K_{c} is given as:

K_{c}=\frac{[NH_3]^2}{[N_2][H_2]^3}

We are given:

[NH_3]=0.25M

[H_2]=0.75M

[N_2]=1.1M

Putting values in above equation, we get:

K_c=\frac{(0.25)^2}{1.1\times (0.75)^3}

K_c=0.135

There are 3 conditions:

  • When K_{c}>1; the reaction is product favored.
  • When K_{c}; the reaction is reactant favored.
  • When K_{c}=1; the reaction is in equilibrium.

For the given reaction, the value of K_c is less than 1. Thus, the reaction is reactant favored.

Hence, the value of K_c is 0.136 and is reactant favored.

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Question 4 (1 point)
Diano4ka-milaya [45]

In an endothermic reaction products are <u>HIGHER </u>than reactants in potential energy and <u>LESS </u>stable.

Explanation:

Energy is input into the reaction in an endothermic reaction. This means the products are of a higher energy level than the reactants. Therefore the reaction increases Gibb's free energy and reduces entropy. Remember in thermodynamic stability involves an increase in entropy and a decrease in Gibbs free energy. Therefore the products are less stable than the reactants. This is why endothermic reactions do not occur spontaneously like exothermic reactions.

6 0
3 years ago
What structure what structural feature of cells can allow molecules to flow directly from the inside of one cell to the inside o
Rufina [12.5K]
The answer is A. Plasmodesmata. It's a narrow pathway between cells.
3 0
3 years ago
3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 1
nordsb [41]

<u>Answer:</u> The enthalpy change of the unknown solid is 3.824 kJ/mol

<u>Explanation:</u>

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

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.20 g/mL

Volume of solution = 150.0 mL

Putting values in above equation, we get:

1.20g/mL=\frac{\text{Mass of solution}}{150.0mL}\\\\\text{Mass of solution}=(1.20g/mL\times 150.0mL)=180g

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T

q = heat absorbed or released

m = mass of solution = 180 g

c =  specific heat capacity = 4.18 J/g°C

\Delta T = change in temperature = 16.01°C

Putting values in above equation:

Q=180\times 4.18\times 16.01=12045.9J=12.046kJ

To calculate the enthalpy change of the reaction, we use the equation

\Delta H_{rxn}=\frac{q}{n}

where,

q = amount of heat absorbed = 12.046 kJ

n = number of moles of solid = 3.15 moles

\Delta H_{rxn} = enthalpy change of the reaction

Putting values in above equation, we get:

\Delta H_{rxn}=\frac{12.046kJ}{3.15mol}=3.824kJ/mol

Hence, the enthalpy change of the unknown solid is 3.824 kJ/mol

5 0
3 years ago
A block of gold is illuminated with UV light at a wavelength of 1.6x10-11 meters. What is the frequency of this light ?
irga5000 [103]

Answer:

f = 1.87 × 10¹⁹ s⁻¹

Explanation:

Given data:

Wavelength of light = 1.6 ×10⁻¹¹ m

Frequency of light = ?

Solution:

Formula:

speed of light = wavelength × frequency

Frequency = speed of light / wavelength

speed of light = 3× 10⁸ m/s

Now we will put the values in formula:

Frequency = speed of light / wavelength

f =  3× 10⁸ m/s / 1.6 ×10⁻¹¹ m

f = 1.87 × 10¹⁹ s⁻¹

4 0
4 years ago
How to balance this equation: <br> Cu(s) + O2(g) --&gt; CuO(s)
Lelu [443]
2Cu(s) + O2(g) --> 2CuO(s)
3 0
3 years ago
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