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kodGreya [7K]
2 years ago
13

Balance the following equations 1.Na+HCI=NaCI+H22.Mg+O2=MgO​

Chemistry
1 answer:
liberstina [14]2 years ago
7 0

Answer:

1. 2Na + 2HCl = 2NaCl =H2

2. 2Mg + O2 = 2MgO

Explanation:

make sure that the number of atoms of each element is the same on each side of the equation  

1. 2Na + 2HCl = 2NaCl =H2

<em><u>on LHS of equation,</u></em>

Na = 2

H = 2

Cl =2

<em><u>On RHS of equation, </u></em>

Na=2

H= 2

Cl =2

2. 2Mg + O2 = 2MgO

<em><u>on LHS of equation,</u></em>

Mg = 2

O2 = 2

<em><u>On RHS of equation, </u></em>

Mg = 2

O2 = 2

<em><u /></em>

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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,

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The intermediate balanced chemical reaction will be,

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(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

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