At equilibirium,1.0 liter vessel was found to contain 0.2moles,0.2moles B,0.4 moles C,and 0.4 moles D.İf 0.1 moles A and 0.1moles B added to system what will be the new equilibirium concentration of A?
A+B=C+D
reversable reaction

Question

24-03-2020
Chemistry

contestada

At equilibirium,1.0 liter vessel was found to contain 0.2moles,0.2moles B,0.4 moles C,and 0.4 moles D.İf 0.1 moles A and 0.1moles B added to system what will be the new equilibirium concentration of A?
A+B=C+D
reversable reaction

Respuesta :

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drpelezo drpelezo · Answer 1 · 2020-03-24T11:53:07+01:00

Answer:

See Explanation and Attachments => New concentration A = 0.10 mol/L

Explanation:

Explanation:

A + B ⇄ C + D

C(i): 0.20 mol 0.20 mol 0.4 mol 0.4 mol

ΔC: -0.10 mol -0.1 mol +0.1 mol +0.1 mol

C(eq): 0.10 mol 0.10 mol 0.5 mol 0.5 mol

__________________

=> A & B react to form more C & D as reaction shifts right to establish new equilibrium with new concentration values. See attached diagram. => I call this the 'chemical see-saw analysis'.

=> Also attached is a summary of stress factor effects of temperature and pressure-volume changes.

At equilibirium,1.0 liter vessel was found to contain 0.2moles,0.2moles B,0.4 moles C,and 0.4 moles D.İf 0.1 moles A and 0.1moles B added to system what will be the new equilibirium concentration of A?A+B=C+Dreversable reaction​

Respuesta :

Otras preguntas

At equilibirium,1.0 liter vessel was found to contain 0.2moles,0.2moles B,0.4 moles C,and 0.4 moles D.İf 0.1 moles A and 0.1moles B added to system what will be the new equilibirium concentration of A?
A+B=C+D
reversable reaction