Question

Consider the following equilibrium N2(g) +3H2(g) 2NH3(g) ΔG--34. kJ Now suppose a reaction vessel is filled with 8.06 atm of nitrogen (N2) and 6.48 atm of ammonia (NH3) at 176. °C. Answer the following questions about this system: rise fall Under these conditions, will the pressure of N2 tend to rise or fall? Dio Is it possible to reverse this tendency by adding H? In other words, if you said the pressure of N2 will tend to rise, can that be changed to a tendency to fall by adding H2? Similarly, if you said the pressure of N2 will tend to fall, can that be changed to a tendency to rise by adding H2 If you said the tendency can be reversed in the second question, calculate the minimum pressure of H2 needed to reverse it. Round your answer to 2 significant digits O yes no atm

Answer 1

as there is no availability of H2, backward reaction takes place.

so that, pressure of N2 will rise.

yes, it is possible to reverse this tendency by adding H2.

DG0 = - RTlnK

-34 = - 8.314*10^-3*(176+273.15)lnKp

Kp = equilibrium constant = 8.99*10^3

Kp = [NH3]^2/[N2][H2]^3

8.99*10^3 = ((6.48-2x)^2/((8.06+x)(3x)^3))

x = 0.0276

pH2 = 3*0.0276 = 0.083 atm

the minimum pressure of H2 required = 0.083 atm