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Tresset [83]
3 years ago
5

Determine the total pressure of a gas mixture that contains O2, N2 and He, if the partial pressure of the gases are : P(O2) = 20

0 mm Hg, P (N2) = 350 mm Hg, P ( He2) = 428 mm Hg
​
Chemistry
1 answer:
natulia [17]3 years ago
5 0

Answer:

P(total) = 978 mmHg

Explanation:

Given data:

Total pressure of gas = ?

Partial pressure of O₂ gas = 200 mmHg

Partial pressure of N₂ gas = 350 mmHg

Partial pressure of He gas = 428 mmHg

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

P(total) = P(O₂) + P(N₂) + P(He)

P(total) = 200 mmHg + 350 mmHg + 428 mmHg

P(total) = 978 mmHg

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

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

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

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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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