1 H2S(g) 2 H2(g) + S2(g) Kp= 9.30x10-8 at 70 degrees celcius. If 0.45 mol of H2S...

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1 H2S(g) 2 H2(g) + S2(g) Kp= 9.30x10-8 at 70 degrees celcius. If 0.45 mol of H2S...

1 H₂S(g)2 H_2(g) + S₂_(g)

Kp= 9.30x10^-8 at 70 degrees celcius. If 0.45 mol of H₂S is placed in a 3.00 L container, what is the equilibrium concentration of H₂_(g) at 700 degrees celcius?

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Answer 1

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(0.45 mol) (0.0821 L. atm/K.mol) (700° C+273)K 3.00 L Initial pressure of H,S = 12.0 atm The ICE table constructed is as foll

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1 H2S(g) 2 H2(g) + S2(g) Kp= 9.30x10-8 at 70 degrees celcius. If 0.45 mol of H2S...

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1 H2S(g) 2 H2(g) + S2(g) Kp= 9.30x10-8 at 70 degrees celcius. If 0.45 mol of H2S...

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1 H2S(g) 2 H2(g) + S2(g) Kp= 9.30x10-8 at 70 degrees celcius. If 0.45 mol of H2S...