Need help with this question using dimensional analysis

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

How many total atoms of each element are presented in the following formula

Scorpion4ik [409]

Answer:

Aluminium (Al): (3*2)+(5*2)=16

Sulphor (S): (3*1)=3

Oxygen (O): (4*3)+(3*1)=15

7 0

3 years ago

How can I differentiate between Ionic compounds, molecular compounds, and acids given only the formula? I need to be able to fin

VikaD [51]

You can differentiate between ionic, covalent and molecular compounds by the fact that ionic compounds contain elements that include both a metal and a nonmetal. Molecular compounds contain both non metals covalently bonded to each other. While acids most often on their chemical formula start with the element of Hydrogen - H.

For Eg - sulphuric acid
H2SO4
hydrochloric acid - HCl.

7 0

3 years ago

calculate the mass of 120cc nitrogen present at STP. how many number of molecules are present in it?

Stells [14]

Answer:

$0.15008\ \text{g}$

$3.23\times 10^{21}$

Explanation:

1 mol of nitrogen at STP = 22.4 L = 22400 cc

n = Mol of $N_2$ = $\dfrac{120}{22400}=0.00536\ \text{mol}$

M = Molar mass of $N_2$ = $28\ \text{g/mol}$

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

Mass of $N_2$ is

$m=nM\\\Rightarrow m=0.00536\times 28\\\Rightarrow m=0.15008\ \text{g}$

Mass of the nitrogen is $0.15008\ \text{g}$

Number of molecules is given by

$nN_A=0.00536\times 6.022\times 10^{23}=3.23\times 10^{21}\ \text{molecules}$

The number of molecules present in it are $3.23\times 10^{21}$

5 0

3 years ago

using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius

lukranit [14]

Answer:

Please, see attached two figures:

The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

Answer: 5.5g

Explanation:

The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the solubility of NH₄Cl.

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

$10mL\times 1.0g/mL=10g$

Thus, the solutibily is:

$10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl$

5 0

3 years ago