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AfilCa [17]
3 years ago
6

Use Hess's law and the following equations to calculate the ΔHreaction for the reaction CO(g) + 3H2(g) CH4(g) + H2O(g). (Show yo

ur work.) (4 points) • C(s) + O2(g) CO(g) ΔH = –110.5 kJ • C(s) + 2H2(g) CH4(g) ΔH = –74.85 kJ • H2(g) + O2(g) H2O(g) ΔH = –241.83 kJ
Chemistry
1 answer:
Vesna [10]3 years ago
6 0

Answer: The enthalpy of reaction is, -206.18 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 final reaction is

CO(g)+3H_2(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=-110.5kJ

(2) C(s)+2H_2(g)\rightarrow CH_4(g)    \Delta H_2=-74.85kJ

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)    \Delta H_3=-241.83kJ

Now adding (2) and (3) and subtracting (1) , we get :

\Delta H_{rxn}=(\Delta H_2+\Delta H_3)-\Delta H_1

\Delta H_{rxn}=((-74.85)+(-241.83))-(-110.5)

\Delta H_{rxn}=-206.18kJ

Therefore, the enthalpy of reaction is, -206.18 kJ

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A sample of nitrogen gas has a temperature of 22.7oC when the volume of the container is 12.2L and it is under 150.4kPa of press
Marrrta [24]

Answer:

The final temperature of the gas would need to be approximately 158.4 K

Explanation:

The details of the sample of nitrogen gas are;

The initial temperature of the nitrogen gas, T₁ = 22.7°C = 295.85 K

The initial volume occupied by the gas, V₁ = 12.2 L

The initial pressure of the gas, P₁ = 150.4 kPa

The final volume of the gas, V₂ = 9.7 L

The final pressure of the gas, P₂ = 101.3 kPa

Let 'T₂', represent the final temperature of the gas, by the ideal gas equation, we have;

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\therefore \ T_2 = \dfrac{P_2 \times V_2 \times T_1 }{P_1 \times V_1}

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\therefore \ T_2 = \dfrac{101.3 \, kPa \times 9.7 \ L \times 295.85  K}{150.4 \ kPa \times 12.2 \ L} \approx 158.4327959 \  K

The final temperature of the gas, T₂ ≈ 158.4 K

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2 years ago
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