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Inessa05 [86]
3 years ago
9

Explain its significance.Check all that apply. Systems with high potential energy tend to change in the direction of lower poten

tial energy, obtaining energy from another objects. Systems with high potential energy tend to change in the direction of lower potential energy, releasing energy into the surroundings. In the exchanges, the total energy is always conserved; energy is neither created nor destroyed. In the exchanges, the total energy is always changed; energy is created and destroyed. In chemical and physical changes, matter never exchanges energy with its surroundings. In chemical and physical changes, matter often exchanges energy with its surroundings.
Chemistry
1 answer:
KatRina [158]3 years ago
5 0

Answer:

your question is not clearly stated, let me assume your question to be ;

Explain the significance of the law of conservation of energy. Check all that apply below.

1. Systems with high potential energy tend to change in the direction of lower potential energy, obtaining energy from another objects. Systems with high potential energy tend to change in the direction of lower potential energy, releasing energy into the surroundings.

2. In the exchanges, the total energy is always conserved; energy is neither created nor destroyed. In the exchanges, the total energy is always changed; energy is created and destroyed.

3. In chemical and physical changes, matter never exchanges energy with its surroundings. In chemical and physical changes, matter often exchanges energy with its surroundings.

Explanation:

1. System with high potential energy tends to change in a direction to system with low potential energy. This system with high potential energy moving towards the direction of lower potential energy will give out their potential energy to the surrounding in form of heat, as they also receive kinetic energy from the surrounding. This is a law of conservation of energy. Energy from the system was transformed from one form to another.

2. Total energy in an exchange is always conserved or transformed.

It is conserved when energy is not lost( an enclosed system). It is changed or termed destroyed when energy is lost to the surrounding in form of heat ( an open system). Energy can neither be created not destroyed but can be transformed from one form to another. This means that the energy that we think was created are the potential energy in the system, while the energy we think was destroyed, are those energy given to the surrounding in form of heat.

3. In a chemical or physical change where matter never exchange energy with it surrounding are called closed system. The exchange of energy in this type of system is only wetin the system.

In a chemical or physical change where matter exchanges energy with it's surrounding are called open system. It is either an Exothermic reaction(the system gives out energy to the surrounding) or Endothermic reaction( the system takes energy from the surrounding).

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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

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Answer:

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Answer:

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