What is the weight in grams of 1 mole of ammonium sulfate (NH4)2SO4

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

Balance the following equation: H20 --> H2 + O2

MakcuM [25]

2H2O = 2H2 + O2.

<h3>Explanation:</h3>

Balancing equations is very essential because of the fact that it represents the stoichiometric quantities of the reactants needed to react to form the product. The ratio of the weights of reactant and product are also very well understood from this.

Here in this equation, the water is broken into hydrogen and oxygen. The balanced reaction is

2H2O = 2H2 + O2.

Two moles of water is broken down into 2 moles of hydrogen and one mole of oxygen.

5 0

3 years ago

Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. cons

Dmitriy789 [7]

Lithium sulfate is Li2SO4

Strontium chlorate is Sr(ClO3)2

Strontium sulfate is SrSO4

So the complete balanced chemical reaction for this is:

Li2SO4 (aq) + Sr(ClO3)2 (aq) --> SrSO4 (s) + 2 LiClO3 (aq)

This is a type of double replacement reaction since there is an exchange of ions.

4 0

3 years ago

Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so

Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:

C₆H₅-COOH + HCO₃⁻ ⇄ C₆H₅-COO⁻ + H₂CO₃

C₆H₅-OH + HCO₃⁻ ⇄ no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl ⇄ C₆H₅-COOH + Cl⁻

This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration.

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻ ⇄ C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl ⇄ C₆H₅-OH + Cl⁻

The precipitated phenol can be isolated by filtration.

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.

I hope it helps you!

6 0

3 years ago

If 4000 g of Fe2O3 reacts, how many moles of CO are needed? (add work)

7nadin3 [17]

Answer:

75.15 mol.

Explanation:

Firstly, we need to write the balanced equation of the reaction:

Fe₂O₃ + 3CO → 2Fe + 3CO₂.

It is clear that 1.0 mole of Fe₂O₃ reacts with 3.0 moles of CO to produce 2.0 moles of Fe and 3.0 moles of CO₂.

∴ Fe₂O₃ reacts with CO with (1: 3) molar ratio.

we need to calculate the no. of moles of (4000 g) of Fe₂O₃:

no. of moles of Fe₂O₃ = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Using cross multiplication:

1.0 mole of Fe₂O₃ needs → 3.0 moles of CO,

∴ 25.05 mole of Fe₂O₃ needs → ??? moles of CO.

∴ The no. of moles of CO needed = (3.0 mol)(25.05 mol)/(1.0 mol) = 75.15 mol.

5 0

4 years ago