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mina [271]
3 years ago
8

For a particular reaction at 235.8 °C, ΔG=−936.92 kJ/mol , and ΔS=513.79 J/(mol⋅K) . Calculate ΔG for this reaction at −9.9 °C.

Chemistry
1 answer:
Rudik [331]3 years ago
8 0

Answer:

-138.9 kJ/mol

Explanation:

Step 1: Convert 235.8°C to the Kelvin scale

We will use the following expression.

K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K

Step 2: Calculate the standard enthalpy of reaction (ΔH°)

We will use the following expression.

ΔG° = ΔH° - T.ΔS°

ΔH° = ΔG° / T.ΔS°

ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K

ΔH° = -3.583 kJ (for 1 mole of balanced reaction)

Step 3: Convert -9.9°C to the Kelvin scale

K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K

Step 4: Calculate ΔG° at 263.3 K

ΔG° = ΔH° - T.ΔS°

ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K

ΔG° = -138.9 kJ/mol

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Easy question, need help
jeka94

Answer: B= 210 amps

Explanation:

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3 years ago
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A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
makvit [3.9K]

Considering the Charles' law, the gas would have a temperature of -109.2 C.

<h3>Charles' law</h3>

Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.

Charles' law is expressed mathematically as:

\frac{V}{T} =k

If you want to study two different states, an initial state 1 and a final state 2, the following is true:

\frac{V1}{T1} =\frac{V2}{T2}

<h3>Temperature of the gas in this case</h3>

In this case, you know:

  • P1= 1800 psi
  • V1= 10 L
  • T1= 20 C= 293 K (being 0 C= 273 K)
  • P2= 1800 psi
  • V2= 6 L
  • T2= ?

You can see that the pressure remains constant, so you can apply Charles's law.

Replacing in the Charles's law:

\frac{10 L}{293 K} =\frac{6 L}{T2}

Solving:

\frac{10 L}{293 K} T2=6 L

T2=\frac{6 L}{\frac{10 L}{293 K} }

<u><em>T2=163.8 K= -109.2 C</em></u>

The gas would have a temperature of -109.2 C.

Learn more about Charles's law:

brainly.com/question/4147359?referrer=searchResults

7 0
3 years ago
If a gas had a volume of 6.7 L and started at STP, what would the new pressure be
Phantasy [73]

The new pressure would be = 4.46 atm

<h3>Further explanation</h3>

Given

V₁=6.7 L(at STP, 1 atm 273 K)

V₂=1.5 L

Required

The new pressure

Solution

Boyle's Law  

At a constant temperature, the gas volume is inversely proportional to the pressure applied  

\rm p_1V_1=p_2.V_2\\\\\dfrac{p_1}{p_2}=\dfrac{V_2}{V_1}

P₂ = (P₁V₁)/V₂

P₂ = (1 atm x 6.7 L)/1.5 L

P₂ = 4.46 atm

5 0
2 years ago
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