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Question
Derive a general expression for the equilibrium constant Kp and Kc for the reaction, \[\ce{3H2(g) + N2(g) <=> 2NH3(g)}\].
Solution
Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and h’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3 x moles of hydrogen to give 2x moles of ammonia.
\[\ce{3H2(g) + N2(g) <=> 2NH3(g)}\]
| N2 | H2 | NH3 | |
| Initial number of moles | a | b | 0 |
| number of moles reacted | X | 3x | 0 |
| Number of moles at equilibrium | a -x | b - 3x | 2x |
| Active mass or molar concentration at equilibrium | `("a -x")/"V"` | `("b" - "3x")/"V"` | `(2x)/"V"` |
Applying law of mass action,
K2 = `(["NH"_3]^2)/(["N"_2]["H"_2]^3)`
= `("2x"/"V")^2/((("a - x")/"V")(("b" - 3"x")/"V")^3)`
= `("4x"^2/"V"^2)/((("a - x")/"V")(("b" - 3"x")/"V")^3)`
= `(4x^2"V"^2)/(("a - x")("b - 3x")^3)`
The equilibrium constant KP can also be calculated as follows:
`"K"_"P" = "K"_"C" ("RT")^(Delta"ng")`
Δng = np - nr = 2 - 4 = - 2
`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["RT"]^-2`
Total number of moles at equilibrium,
n = a - x + b - 3x + 2x = a + b - 2x
`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["PV"/"n"]^-2`
`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx ["n"/"PV"]^2`
`"K"_"P" = (4x^2"V"^2)/(("a - x")("b - 3x")^3) xx [("a + b - 2x")/"PV"]^2`
`"K"_"P" = (4x^2("a + b - 2x")^2)/("P"^2("a - x")("b - 3x")^3)`
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