Ag could be eroded by gaseous H_2S via the following reaction: H_2S(g) + 2Ag(s)
ID: 897300 • Letter: A
Question
Ag could be eroded by gaseous H_2S via the following reaction: H_2S(g) + 2Ag(s) rightarrow Ag_2S(s) + H_2(g) It is known that at 298 K, 1 atm, the standard Gibbs energy of formation of Ag_2S(s) and H_2S(g) are -40.26 kJ mol^-1 and -33.02 kJ mol^-1, respectively. At 298K and 1 atm, please answer: Will Ag be eroded to form Ag_2S(s) if exposed to a mixture of H_2S(g) and H_2(g) with 1:1 volume ratio? What is the maximum mole fraction of H_2S(g) in H_2(g)&H;_2S(g) mixture that can ensure Ag will not be eroded?Explanation / Answer
A)
1st calculate delta G of rxn
delta Go(rxn) =delta Gf (H2) + delta Gf(Ag2S) - delta Gf(H2S)-2*delta Gf(Ag)
= 0 +(-40.26)- (-33.02) -0
= -7.24 KJ
= -7240 J
use:
delta G = delta Go-R*T*ln Kc
delta G = delta Go + R*T*ln {[H2]/[H2S]}
delta G = delta Go + R*T*ln {1/1}
delta G = delta Go + 0
= -7240 J
B)
delta G = delta Go + R*T*ln {[H2]/[H2S]}
delta G>0 for not eroding
delta Go + R*T*ln {[H2]/[H2S]} > 0
-7240 +8.314*298 * ln {[H2]/[H2S]}
ln {[H2]/[H2S]}>2.92
[H2]/[H2S]>18.58
Maximum H2S 1/18.58 = 0.054
Answer:0.054