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in a mixture if N2 and H2 in the reaction 1:3 at 30atm and 300°C , the% of NH3 at equilibrium is 17.8.Calculate kp for N2+H2=2NH3 |
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Answer» Answer: What would you like to ASK? 11th Chemistry Equilibrium Solubility Equilibria In a mixture of N2 and H2 i... CHEMISTRY In a mixture of N 2
and H 2
initially in a mole ration of 1:3 at 30 atm and 300 o C, the percentage of ammonia by volume under the equilibrium is 17.8. CALCULATE the equilibrium constant (K P
) of the mixture, for the reaction, N 2
(g)+3H 2
(g)⇌2NH 3
(g) HARD Share Study later ANSWER Let the initial moles of N 2
and H 2
be 1 and 3. N 2
+3H 2
⟶2NH 3
Initial 1 3 0 at Eqb. 1−x 3−3x 2X % of volume is same as % by mole NH 3
4−2x 2x
=0.178 x= 2+(2×0.178) 4×0.178
x=0.302 mole fraction of H 2
at eqb.= 4−2x 3−3x
=0.6165 Mole fraction of N 2
at eqb.=1−0.6165=0.178=0.2055 K p
=(X NH 3
.P T
) 2 /(X N 2
)(X H 2
.P T
) 3
= (0.2055×30)(0.6165×30) 3
(0.178×30) 2
=7.31×10 −4 atm −2 |
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