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Chemical Reactivity

N_2 andH_2 ar...

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N_{2 }andH_2 are taken in 1:3molar ratio. ATequilibrium total pressure is P and NH_{3 }is 40%by mole. then k_p for N_2+{H_{2=}}_{}NH_{3 }is

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45^{0} C

the equilibrium constant

K_{P}

for the reaction

N_{2} + {3 H}_{2} ⇌ {2 N H}_{3}

was found to be

1.6 \times 10^{- 5}

at a pressure of 200 atm. If

N_{2}

H_{2}

are taken in a ratio 1:3. What is the % of

{N H}_{3}

formed at this temperature?

Q. In a mixture of

N_{2}

H_{2}

in the ratio 1:3 at 30 atm and

300^{0} C

{N H}_{3}

at equilibrium is 17.8.
Calculate

K_{P}

{N_{2} + {3 H}_{2} ⇌ 2 N H}_{3}

.

The equilibrium constant $K_{p}$ for the reaction $N_{2}$ (g) + 3 $H_{2}$ (g) $⇋$ $2 N H_{3}$ (g) is:

450^{o} C

, the equilibrium constant

K_{p}

for the reaction,

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

, was found to be

1.6 \times 10^{- 5}

at a pressure of

200 a t m

N_{2}

H_{2}

1 : 3

ratio, what is % of

{N H}_{3}

formed at this temperature?

Q. A reaction is carried out by 1 mole of

N_{2}

H_{2}

shows at equilibrium, the mole fraction of

N H_{3}

is 0.012 at 500

^{o} C

and 10 atm pressure.

K_{p}

is the equilibrium constant. The pressure (in atm) at which mole % of

N H_{3}

in equilibrium mixture increased to 10.4 is_________.
[Give your answer in closest integer]

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