I need help, Asap!!!!!! this is for tomorrow

Anna [14]

Considering the Hess's Law, the enthalpy change for the reaction is 221.8 kJ/mol.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol

Equation 2: C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol

Equation 3: 2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

<h3 /><h3>FIRST STEP</h3>

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C₂H₄ (g) on reactant side and it is present in first equation. Since this equation has one mole of C₂H₄ (g) on the product side, it is necessary to locate it on the reactant side (invert it).

When an equation is inverted, the sign of ΔH° also changes.

<h3>SECOND STEP</h3>

Now, you need 2 moles of CF₄ (g) on the product side. The second equation has 1 mole of CF₄ (g) on the product side, so it is necessary to multiply it by 2 to obtain 2 moles of CF₄ (g).

Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

<h3>THIRD STEP</h3>

Finally, you need 4 moles of HF (g) on the product side. The first equation has 2 moles of HF (g) on the product side, so it is necessary to multiply it by 2 to obtain 4 moles of the compound.

Since the equation is multiply by 2, the variation of enthalpy also is multiplied by 2.

<h3>SUMMARY</h3>

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 H₂ (g) + 2 F₂ (g) → 4 HF (g) ∆H° = -158.4 kJ/mol

Equation 2: 2 C (s) + 4 F₂ (g) → 2 CF₄ (g) ∆H° = 282.6 kJ/mol

Equation 3: C₂H₄ (g) → 2 C(s) + 2 H₂ (g) ∆H° = 97.6 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g) ΔH°= 221.8 kJ/mol

Finally, the enthalpy change for the reaction is 221.8 kJ/mol.

Learn more about molar enthalpy:

7 0

2 years ago

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol. How much energy is produced by burning 0.8210 g

bazaltina [42]

The energy produced by burning : -32.92 kJ

<h3>Further explanation</h3>

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The enthalpy and heat(energy) can be formulated :

$\tt \Delta H=\dfrac{Q}{n}\rightarrow n=mol$

The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol.

The energy released for 0.8210 g of naphthalene :

$\tt Q=\Delta H\times n\\\\Q=-5139.6~kJ/mol\times \dfrac{0.8210~g}{128.17~g/mol}~\\\\Q=-32.92~kJ$

3 0

3 years ago

How many moles of H2 are produced from 5.8 moles of NH3

vodomira [7]

Answer:

1. 8.7moles of H2

2. 2.25moles of O2

Explanation:

1. 2NH3 —> N2 + 3H2

From the equation,

2moles of NH3 produce 3 moles of H2.

Therefore, 5.8moles of NH3 will produce Xmol of H2 i.e

Xmol of H2 = (5.8x3)/2 = 8.7moles

2. C3H8 + 5O2 —> 3CO2 + 4H2O

From the equation,

5moles of O2 produced 4moles of H2O.

Therefore, Xmol of O2 will produce 1.8mol of H2O i.e

Xmol of O2 = (5x1.8)/4 = 2.25moles

4 0

3 years ago

The following reaction has an equilibrium constant (Keq) of 1 under standard conditions and can be catalysed by the enzyme aspar

Trava [24]

Answer:

What is the AGⓇ of this reaction? 0.
Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.

Explanation:

1. To calculate the delta G of a reaction given the K, we use the following equation:

ΔG°= -RT ln K.

Which gives us 0 when K is 1.

2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.

3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.

4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.

4 0

3 years ago

Using the equations and enthalpy values provided, which mathematical expression can be used to determine the unknown enthalpy ch

vlabodo [156]

In order to solve this, we need to make use of Hess' Law.

We are already given the equations and their corresponding deltaH. Using Hess' Law, we can generate this equation:
104 kJ = x - (-1182 kJ) - (-1144 kJ)

Among the choices, the answer is
<span>B.104 = x - [(-1182) + (-1144)]
</span>

3 0

3 years ago

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