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Consider the following reaction: N2(g) + 3H2(g) 2NH3(g) Report all answers to th

ID: 628264 • Letter: C

Question

Consider the following reaction: N2(g) + 3H2(g) 2NH3(g) Report all answers to three significant figures! a) Determine the standard free energy of the the reaction at 298K? Delta G= -33 kJ b) Determine the equilibrium constant at 298K? K= 608950.2 c) Determine the reaction quotient when the partial pressures of the gases are the following. N2 = 10.90 atm; H2 = 14.40 atm; NH3 = 1.00 atm Q= _______ d) Determine the value of G when the reaction mixture is the one described in Part c. Delta G= ______ kJ I need help finding parts c and d! Thanks!

Explanation / Answer

Given that:

N2+3H2=2NH3

So Qp= 1/N2*(H2)^3 i.e. coeff. of H2 in reaction is 3

And substiute values

Qp=1/10.9*14.4^3

Qp=3.07*10^-5 atm^-2

And we know relation

Qc=Qp(RT)^n

Qc=3.07*10^-5(0.0821*298)^2

n= Coeff. of product-Coeff. of reactant

n=2-4=-2

So, Qc= 0.01837

B)Delta g=-RTlnQc= -8.314*298ln(0.01837)

Delta G= 9900 J/mol

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