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For the gaseous reaction, N 2 + 3H 2 ↔ 2NH 3 the partial pressures of H 2 and N 2 are 0.4 and 0.8 atm respectively. The total pressure of the entire system is 2.8 atm. What will be the value of K p if all the concentrations are given in atmospheres??
N2 + 3H2 ↔ 2NH3 partial pressures of H2 and N2 are 0.4 and 0.8 atm so partial pressure of 2NH3 = 1.6 atm Kp= [NH3] ^2 / ( [N2] [ H2]^3 ) = 1.6 ^ 2 / 0.8 * 0.4 ^3 = 2.56 / 0.0512 = 50 Please approve !
N2 + 3H2 ↔ 2NH3
partial pressures of H2 and N2 are 0.4 and 0.8 atm
so partial pressure of 2NH3 = 1.6 atm
Kp= [NH3] ^2 / ( [N2] [ H2]^3 )
= 1.6 ^ 2 / 0.8 * 0.4 ^3
= 2.56 / 0.0512
= 50
Please approve !
if partial pressures tht u hav givn fr h2 nd n2 r at equilibrium thn... presure of nh3=1.6 so kp=(1.6)^2/0.8*(0.4)^3
if partial pressures tht u hav givn fr h2 nd n2 r at equilibrium thn...
presure of nh3=1.6
so kp=(1.6)^2/0.8*(0.4)^3
N2 + 3H2 => 2NH3 0.4 0.8 P (at equilibrium) sum of all pressure = P(total) = 1.2 + p = PT PT = 2.8 (given) so 1.2 + P = 2.8 pNH3 = 1.6 now Kp = ( pNH3)2 /(PN2)(pH2)3 = (1.6)2/(0.4)(0.8)3 Kp = 12.5 (atm)-2
N2 + 3H2 => 2NH3
0.4 0.8 P (at equilibrium)
sum of all pressure = P(total) = 1.2 + p = PT
PT = 2.8 (given) so
1.2 + P = 2.8
pNH3 = 1.6
now Kp = ( pNH3)2 /(PN2)(pH2)3
= (1.6)2/(0.4)(0.8)3
Kp = 12.5 (atm)-2
thnx to al......
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