Answer the following questions using the chemical reaction and thermochemical in

Question

Answer the following questions using the chemical reaction and thermochemical information given below:



NH3(g) + HI(g) NH4I(s)

1. Determine G°rx (in kJ) for this reaction at 412.80 K. Assume H°f and S° do not vary as a function of temperature. Report your answer to two decimal places
Answer:
-94.12 kJ

B/ Determine the equilibrium constant for this reaction at 412.80 K. Report your answer to three significant figures in scientific notation.

C/ If the partial pressure of HI is 4.26 atm and the partial pressure of NH3 is 3.24 atm, determine G (in kJ) for this reaction at 412.8 K. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ)

Hf° (kJ/mol)     S° (J mol-1 K-1) NH4I -201.40 186.90 NH3 -45.94 192.77 HI 26.36 206.59

Explanation / Answer

we know that

dHo rxn = dhfo products - dHfo reactants

so

dHo rxn = dHfo NH4I - dHfo NH3 - dHfo HI

so

dHo rxn = -201.4 + 45.94 - 26.36

dHo rxn = -181.82 kJ / mol

now

dSo rxn = dSo NH4I - dSo NH3 - dSo HI

dso rxn = 186.90 - 192.77 - 206.59

dSo rxn = -212.46 J / mol K

now

we know that

dGo = dHo - (TdSo)

so

dGO = ( -181.82 x 1000) - ( 412.8 x -212.46)

dGo = -94.12 kJ

now

2)

we know that

dGo = -RT ln Keq

so

-94.12 x 1000 = -8.314 x 412.80 x ln Keq

Keq = 8.13 x 10^11

so

the value of equilibrium constant is 8.13 x 10^11

3) now

NH3 (g) + HI (g) ---> NH4I (s)

solids are not consider for equilibrium constant

so

reaction quotient (Q) = 1 / (pNH3) x (pHI)

so

Q = 1 / 3.24 x 4.26

Q = 0.07245

now

we know that

dG = dGo + RT ln Q

so

dG = (-94.12 x 1000) + ( 8.314 x 412.8 x ln 0.07245)

dG = -103.13 kJ

so

the value of dG is -103.13 kJ

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