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kicyunya [14]
3 years ago
6

A rigid stainless steel chamber contains 180 Torr of methane, CH 4 , and excess oxygen, O 2 , at 160.0 °C. A spark is ignited in

side the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.
Chemistry
1 answer:
I am Lyosha [343]3 years ago
7 0

Answer:

No pressure will change in the chamber at reactions end

Explanation:

CH4 + 2O2 ---> CO2 + 2H2O

According the above reaction

There are 3 moles total of gaseous reactants, and 3 moles total of gaseous products, so this reaction results in no change of volume. Therefore, since this happens at constant volume and constant temperature, the pressures are directly related to the moles used and produced. This way, no pressure will change in the chamber at reactions end

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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 Hess’s law, the chemical equation can be treated as algebraic expressions 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.

P_4(s)+6Cl_2\rightarrow 4PCl_3  \Delta H_1=-2439kJ (1)

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Net chemical equation:

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Adding 1 and 2 we get,

4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2 \Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ   (4)

Now dividing equation (4) by 4, we get

PCl_5(g)\rightarrow PCl_3(g)+Cl_2

\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ   (4)

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