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vazorg [7]
3 years ago
13

The equilibrium constant Kc for the reaction

Chemistry
1 answer:
PIT_PIT [208]3 years ago
3 0

Answer:

[N2] = 0.3633M

[H2] = 1.090M

[NH3] = 0.2734M

Explanation:

Based on the reaction of the problem, Kc is defined as:

Kc = 0.159 = [NH3]² / [N2] [H2]³

<em>Where [] are the equilibrium concentrations.</em>

The initial concentrations of the reactants is:

N2 = 1.00mol / 2.00L = 0.500M

H2 = 3.00mol / 2.00L = 1.50M

When the equilibrium is reached, the concentrations are:

[N2] = 0.500M - X

[H2] = 1.50M - 3X

[NH3] = 2X

<em>Where X is reaction quotient</em>

Replacing in the Kc equation:

0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³

0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴

0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²

0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0

Solving for X:

X = 0.1367. Right solution.

X = 1.8286. False solution. Produce negative concentrations

Replacing:

[N2] = 0.500M - 0.1367M

[H2] = 1.50M - 3*0.1367M

[NH3] = 2*0.1367M

The equilibrium concentrations are:

<h3>[N2] = 0.3633M</h3><h3>[H2] = 1.090M</h3><h3>[NH3] = 0.2734M</h3>

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A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to
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Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

<u>Step 2:</u> Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0
3 years ago
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