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Imdadul Haque posted an Question
September 21, 2020 • 00:23 am 30 points
  • IIT JAM
  • Chemistry (CY)

At constant pressure, the presence of argon at the equilibrium of na(g)+3h(g)2nh(8) will (a) reduce the formation of nh3 (b) increase the formation of nh ()redu

At constant pressure, the presence of argon at the equilibrium of Na(g)+3H(g)2NH(8) will (a) reduce the formation of NH3 (b) increase the formation of NH ()reduce the formation of H, (d) increase the formation of both N, and H2

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    Priyanshu kumar Best Answer

    The reaction for the formation of ammonia is N2​(g)+3H2​(g)⇔2NH3​(g). The forward reaction occurs with a decrease in the number of moles from 4 to 2. Also, the reverse reaction occurs with an increase in the number of moles from 2 to 4. When inert gas such as argon is added at constant pressure, the equilibrium will shift in the direction in which there is an increase in the number of moles of gases. Thus, the addition of argon gas at constant pressure will reduce the formation of ammonia from nitrogen and hydrogen.

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    Option A is correct

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    Sir Option D q correct nehii haii ??

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    forward reaction favour karega

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