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hram777 [196]
3 years ago
7

How to draw Hess' Cycle for this question ?

Chemistry
1 answer:
NISA [10]3 years ago
7 0

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

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

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

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

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/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

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muminat

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A, 4Fe 3O2 → 2Fe2O3

5 0
3 years ago
In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.
kirill115 [55]

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}

n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{Fe} = 0.5\:mol}

* to n(S)

n_{S} = \dfrac{m_{S}}{MM_{S}}

n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}

\boxed{n_{S} = 0.5\:mol}

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

Fe + S \Longrightarrow FeS

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}

m_{FeS} = n_{FeS}*MM_{FeS}

m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}

\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}

Answer:

44 grams of iron sulfide

___________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

4 0
3 years ago
3. What role did the government play in either supporting or suppressing African Americans'/Blacks'
9966 [12]
Slavery played a big role in this. the confederacy allowed slavery while the union was oppose to it.
4 0
2 years ago
A sample of a gas in a balloon has a volume of 2.22 L and temperature of 23.9 °C. Calculate the volume when the temperature is r
nignag [31]

Answer:

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Explanation:

Charles's gas law relates the volume and temperature of a gas at constant pressure. This law says that at constant pressures if we raise the temperature of a gas it will expand and if we reduce the temperature the volume will decrease. The formula is as follows:

{\displaystyle {\frac {V_{1}}{T_{1}}}={\frac {V_{2}}{T_{2}}}}

So, the initial conditions are 2.22L and 23.9 °C ant the final conditions are 46.1 °C we replace them in the equation. And then we solve it.

{\displaystyle {\frac {2.22}{23.9}}={\frac {V_{2}}{46.1}}

{\displaystyle {\frac {2.22}{23.9}} {46.1}={V_{2}

{\displaystyle {4.28}={V_{2}

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35. Which of the following atoms has the most possible bond sites?
Vadim26 [7]

Answer:

C phosphorus has 5 bond sites.

4 0
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