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The ammonia in equilibrium with a 1:3 Nirtrogen and hydrogen mixture at 20 atm and 427 degree Celsius amounts to 16%.calculate kp and kc for... 1/2 nitrogen + 3/2 hydrogen reversible Ammonia
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The ammonia in equilibrium with a 1:3 Nirtrogen and hydrogen mixture a...
Calculation of Kp and Kc for the Reaction 1/2N2 + 3/2H2 ⇌ NH3


Equilibrium Constant - Kp

The equilibrium constant Kp can be calculated using the expression:


Kp = (P(NH3))^2 / (P(N2)^(1/2) * P(H2)^3/2)

where P(NH3), P(N2) and P(H2) are the partial pressures of ammonia, nitrogen and hydrogen respectively at equilibrium.


Given that the ammonia amounts to 16% of the equilibrium mixture, the partial pressures of the gases at equilibrium can be determined:


P(NH3) = 0.16 * 20 atm = 3.2 atm
P(N2) = 0.84 * 20 atm = 16.8 atm
P(H2) = 0.84 * 20 atm * (3/2) = 25.2 atm

Substituting these values into the expression for Kp:


Kp = (3.2)^2 / (16.8^(1/2) * 25.2^3/2)
Kp = 1.8 x 10^-4

Equilibrium Constant - Kc

The equilibrium constant Kc can be calculated using the expression:


Kc = ([NH3]^2) / ([N2]^(1/2) * [H2]^3/2)

where [NH3], [N2] and [H2] are the molar concentrations of ammonia, nitrogen and hydrogen respectively at equilibrium.


Since the equilibrium mixture contains only gases, the ideal gas law can be used to relate the partial pressures and molar concentrations of the gases:


P = nRT/V => n/V = P/RT

where P is the partial pressure, n is the number of moles, V is the volume, R is the gas constant and T is the temperature.


Using this relationship, the molar concentrations of the gases can be calculated:


[NH3] = (P(NH3) / RT) = (3.2 atm / 0.0821 L atm/mol K / 700 K) = 0.0052 M
[N2] = (P(N2) / RT) = (16.8 atm / 0.0821 L atm/mol K / 700 K) = 0.0274 M
[H2] = (P(H2) / RT) = (25.2 atm / 0.0821 L atm/mol K / 700 K) = 0.0411 M

Substituting these values into the expression for Kc:


Kc = (0.0052)^2 / (0.0274^(1/2) * 0.0411^3/2)
Kc = 5.1 x 10^-4

Explanation

The equilibrium constant Kp and Kc are measures of the extent of the reaction at equilibrium, and can be used to predict the direction in which the reaction will
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The ammonia in equilibrium with a 1:3 Nirtrogen and hydrogen mixture a...
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A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not define d). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.If four identical samples of an ideal gas initially at similar state (P0, Vo, T0) are allowed to expand to double their volumes by four different processes..I : by isothermal irreversible process II : by reversible process having equation P2V = constant III : by reversible adiabatic process IV: by irreversible adiabatic expansion against constant external pressure.Then in the graph shown the final state is represented by four different points then, the correct match can be

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The ammonia in equilibrium with a 1:3 Nirtrogen and hydrogen mixture at 20 atm and 427 degree Celsius amounts to 16%.calculate kp and kc for... 1/2 nitrogen + 3/2 hydrogen reversible Ammonia
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