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ArbitrLikvidat [17]

2 years ago

12

When the reaction 2H₂S(g) ↔️2H₂(g) + S2(g) is carried out at 1065°C, Kp = 0.012. Starting with pure H₂S at 1065°C, what must the

initial pressure of H₂S be if the equilibrated mixture at this temperature is to contain 0.250 atm of H₂(g)?

1 answer:

mario62 [17]2 years ago

8 0

Answer:

Answer is in the attachment.

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Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

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<u>Answer:</u> The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

<u>Explanation:</u>

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<u>Equation 1:</u> $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

<u>Equation 2:</u> $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

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$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

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$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

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