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pogonyaev
1 year ago
9

Given that E

Chemistry
1 answer:
bija089 [108]1 year ago
5 0

The equilibrium constant of the reaction is 1.21 * 10^6 while the change in free energy is -34.7 kJ.

<h3>What is equilirium constant?</h3>

The equilibrium constant shows the extent of conversion of reactants to products.

Now we know from the Nernst equation that;

Ecell = E°cell - 0.0592/n logQ

E°cell = 0.52−0.16=0.36 V

Since Ecell = 0 V at equilibrium,

0 = 0.36 - 0.0592/1 log K

0.36  = 0.0592/1 log K

log K = 0.36/ 0.0592

K = antiog (0.36/ 0.0592)

K = 1.21 * 10^6

ΔG = -RT lnK

ΔG =-(8.314 * 298 * ln1.21 * 10^6)

ΔG =-34.7 kJ

Learn more about equilibrium constant:brainly.com/question/10038290

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One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K
mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy (s_{2} - s_{1}), in joules per gram-Kelvin, by the following model:

s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }

s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}} (1)

Where:

m - Mass, in kilograms.

c_{w} - Specific heat of water, in joules per kilogram-Kelvin.

T_{1}, T_{2} - Initial and final temperatures of water, in Kelvin.

If we know that m = 1\,kg, c_{w} = 4190\,\frac{J}{kg\cdot K}, T_{1} = 373.15\,K and T_{2} = 288.15\,K, then the change in entropy for the entire process is:

s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}

s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0
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Why is stability of compounds essential
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Answer:

chemical stability is important to consider in the comprehensive assessment of pharmaceutical properties, activity, and selectivity during drug discovery.

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What else is produced during the combustion of butane, C4H10?
denpristay [2]

Another product: CO₂

<h3>Further explanation</h3>

Given

Reaction

2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O

Required

product compound

Solution

In the combustion of hydrocarbons there can be 2 kinds of products

If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product

If Oxygen is low, you'll get Carbon monoxide(CO) and water

Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products

if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left

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