In the following reaction, we start with 2 moles of N2 and 5 moles of H2 exerting a total prssure of 7 aim at a given temperature in a closed vessel. When 50% of N2 s converted into NH3, partial pressure of NH3 is N2+3H2 xrightarrow[]2NH3

Question

In the following reaction, we start with 2 moles of N2 and 5 moles of H2 exerting a total prssure of 7 aim at a given temperature in a closed vessel. When 50% of N2 s converted into NH3, partial pressure of NH3 is N2+3H2 xrightarrow[]2NH3  (A) 2.8 atm (B) 2atm (C) 3.2 atm (D) 4 atm

Tardigrade Admin · Accepted Answer

N2+3H2 xrightarrow[]2NH3 Initial moles 2~~~~~~~~~~5~~~~~~~~~~~~~0 reacted (50% N2 ) ~1~~~~~~~~~3~~~~~~~~~~~~~~2 Left moles. 1~~~~~~~~~2~~~~~~~~~~~~~~2 Total number of moles of N2H2 and NH3 left =1+2+2=5 Initial moles = 7 pressure at initials =7 atm pressure at equilibrium =5 atm Partial pressure of NH3=(2/5)× 5=2 atm

Q. In the following reaction, we start with 2 moles of N2​ and 5 moles of H2​ exerting a total prssure of 7 aim at a given temperature in a closed vessel. When 50% of N2​ s converted into NH3​, partial pressure of NH3​ is N2​+3H2​​2NH3​

2.8 atm

2atm

3.2 atm

4 atm

N2​+3H2​​2NH3​ Initial moles 2 5 0 reacted (50% N2​ ) 1 3 2 Left moles. 1 2 2 Total number of moles of N2​H2​ and NH3​ left =1+2+2=5 Initial moles = 7 pressure at initials =7 atm pressure at equilibrium =5 atm Partial pressure of NH3​=52​×5=2atm

Q. In the following reaction, we start with 2 moles of $N_{2}$ and 5 moles of $H_{2}$ exerting a total prssure of 7 aim at a given temperature in a closed vessel. When 50% of $N_{2}$ s converted into $N H_{3},$ partial pressure of $N H_{3}$ is $N_{2} + 3 H_{2} 2 N H_{3}$