At 90° C, the following equilibrium is established H2(g)+S(s)H2S(g); Kc=6.8× 10-2 If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to 90° C in a 1.0 L vessel, what will be the partial pressure of H2S at equilibrium?

Question

At 90° C, the following equilibrium is established H2(g)+S(s)H2S(g); Kc=6.8× 10-2 If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to 90° C in a 1.0 L vessel, what will be the partial pressure of H2S at equilibrium? (A) 0.25 atm (B) 0.38 atm (C) 0.46 atm (D) 0.56 atm

Tardigrade Admin · Accepted Answer

H2(g)+S(s)H2S(g); Kc=6.8× 10-2 Initial 0.20 0 Atequiation (0.20-x)~ x Kc=6.8× 10-2=([H2S]/H2) 6.8× 10-2=(x/0.20-x) x=1.273× 10-2mol L-1 pH2S=( (n/v) )RT =(1.273× 10-2)× 0.0821× 363 =0.38 text atm

Q. At 90∘C, the following equilibrium is established H2​(g)+S(s)H2​S(g); Kc​=6.8×10−2 If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to 90∘C in a 1.0 L vessel, what will be the partial pressure of H2​S at equilibrium?

0.25 atm

0.38 atm

0.46 atm

0.56 atm

H2​(g)+S(s)H2​S(g); Kc​=6.8×10−2 Initial 0.20 0 Atequiation (0.20−x) x Kc​=6.8×10−2=H2​[H2​S]​ 6.8×10−2=0.20−xx​ x=1.273×10−2molL−1 pH2​S​=(vn​)RT =(1.273×10−2)×0.0821×363 =0.38atm

Q. At $90^{\circ} C,$ the following equilibrium is established $H_{2} (g) + S (s) H_{2} S (g);$ $K_{c} = 6.8 \times 10^{- 2}$ If 0.20 mole hydrogen and 1.0 mole of sulphur are heated to $90^{\circ} C$ in a 1.0 L vessel, what will be the partial pressure of $H_{2} S$ at equilibrium?