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Kipish [7]
2 years ago
9

The internal energy of a system is always increased by ________. A. adding heat to the system B. adding heat to the system and h

aving the system do work on the surroundings C. withdrawing heat from the system D. having the system do work on the surroundings
Chemistry
2 answers:
spin [16.1K]2 years ago
7 0

Answer:

C- withdrawing heat from the system

Explanation:

Internal energy (U) is defined as the total energy of a closed system. Internal energy is the sum of potential energy of the system and the system's kinetic energy and the internal energy of a system can be increased by introduction of matter, by heat, or by doing thermodynamic work on the system.

lord [1]2 years ago
3 0

Answer:

B. adding heat to the system and having the system do work on the surroundings

Explanation:

The internal energy of a system is the energy contained within the system. From first law of thermodynamics we have the equation : dq=du+dw

and we know that energy can neither be created nor destroyed; energy can only be transferred or changed from one form to another therefore du is zero. dq = dw this means that the entire heat supplied is converted into work (on the surroundings)

However, some of the heat supplied is also used to increase the internal energy of the system

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The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
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K=A\times e^{\frac{-Ea}{RT}}

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\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}

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K_1 = rate of reaction without catalyst

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Ea_1 = activation energy without catalyst  = 184 kJ/mol = 184000 J/mol

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Now put all the given values in this formula, we get:

\frac{K_2}{K_1}=e^{\frac{184,000 kJ-59000 kJ}{R\times 300}}=7.632\times 10^{10}

The reaction enhances by 7.632\times 10^{10}  when catalyst is present.

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The catalyzed reaction will take time of 5.11\times 10^{-8} years.

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