∆G° for the reaction of nitrogen with hydrogen to produce ammonia (see balanced chemical equation below) has a value of -33.3 kJ/mol. What will be ∆G for this reaction at 25° C if the reaction is initiated with 125 atm each of N2 and H2 and 2.00 atm NH3(g)? N2(g) + 3H2(g) 2NH3(g)

-0.85KJ

Explanation:

Given N2(g) + H2(g) <--->2NH3(g)

Kp =[ P(NH3)]²/[P(H2)]³[P(N2)]

Where P is the pressure of the gas

P(H2)b= P(N2) = 125atm

P(NH3) = 200atm

Kp = 2²/(125)³(125)

Kp = 2.048 ×10^-6

∆G = -RTlnKp

R =0.008314 J/Kmol

T = 25 +273/= 298k

= 8.314 ×10^-3 × 298 × ln(2.048 ×10^-6)

= -0.008314 × 298 × (-13.099)

= 32.45KJ

∆G = ∆G° + RTlnKp

∆G = -33.3 + 32.45

∆G = -0.85KJ or -850J

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.

the answer is c

tell me if wrong