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Explain the effect on value of Qc by dec...

Explain the effect on value of `Q_c` by decrease the volume of `CO_((g)) + 3H_(2(g)) hArr CH_(4(g)) + H_2O_((g))` reaction vessel.

Text Solution

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If the volume is decrease so pressure is increase and gaseous mol decrease (2 out of 4) therefore reaction is forward. To increase the pressure, the partial pressure of reactions `([CO] + [H_2])_((g))` increase and it is in dominator in the formuly of `Q_c` so `Q_c` decreases.
`Q_c=([CH_4][H_2O])/([CO][H_2]^3) gt K`
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