Can u pls explain me the variation of total pressure of a reaction with time like decomposition of ethylene oxide into CH4 and CO
Dear Student,
Please find below the solution to the asked query:
For the decomposition of ethylene oxide into methane and carbon monoxide, the reaction is
CH2CH2O CH4 + CO
1 mole of ethylene oxide decomposes to form 1 mole each of methane and carbon monoxide. So, 1 mole of reactant gives rise to 2 moles of products.
According to the ideal gas equation, at constant temperature and volume, the pressure of a gas is directly proportional to the number of moles of the gas. As moles of gas increases as reaction proceeds, the total pressure also increases.
For a reaction that has lesser number of moles of products than reactants, the total pressure of the reaction decreases with progress of the reaction. An example of such reaction is, N2 + 3H2 2NH3
Hope this information will clear your doubts about change of total pressure with time for a reaction involving gases.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
Please find below the solution to the asked query:
For the decomposition of ethylene oxide into methane and carbon monoxide, the reaction is
CH2CH2O CH4 + CO
1 mole of ethylene oxide decomposes to form 1 mole each of methane and carbon monoxide. So, 1 mole of reactant gives rise to 2 moles of products.
According to the ideal gas equation, at constant temperature and volume, the pressure of a gas is directly proportional to the number of moles of the gas. As moles of gas increases as reaction proceeds, the total pressure also increases.
For a reaction that has lesser number of moles of products than reactants, the total pressure of the reaction decreases with progress of the reaction. An example of such reaction is, N2 + 3H2 2NH3
Hope this information will clear your doubts about change of total pressure with time for a reaction involving gases.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards