Question

A sample of H2 with a pressure of 1.01 atm and a volume of 210 mL is allowed to react with excess N2 at 141 °C.

N2 (g) + 3 H2 (g) 2 NH3 (g)

Calculate the pressure of the NH3 produced in the reaction if it is transferred to a 1.54-L flask and cooled to 33 °C. Answer should be in atm.

Answer 1

0.068atm

Explanation

No of moles H2 is obtained by Ideal gas equation

PV = nRT

n =PV/RT

= 1.01atm×0.210L/(0.082057L atm /mol K ×414.15K)

= 0.006241mol

N2(g) + 3H2(g) -------> 2NH3

From balanced equation we know that 2 moles of NH3 is formed from 3moles of H2

Therefore,

0.006241moles of H2 will give (2/3)×0.006241=0.004161moles of NH3

pressure of NH3 is calculated from ideal gas equation

PV = nRT

P = nRT/V

= (0.004161mol × 0.082057L atm /mol K × 306.15K)/1.54L = 0.068atm