At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 .

If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

Question

At 90&deg;C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 &times; 10 ^2 . If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90&deg;C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

Arpit Jaiswal · Answer

given, H2 (g) + S(s) = H2 S(g) Kp = 6.8 &times; 10 ^2 int. moles 0.2 1.0 ---- final moles (0.2-x) (1.0-x) x total moles are 1.2-x therefore, from Kp equation we have- 6.8 x 10^2 = [x/(1.2-x) * 1] [(0.2-x)/(1.2-x)*1][(1-x)/(1.2-x)*1] solving the above quadratic, we get- x now partial pressure of H2 S is (x/1.2-x)*1 which is the required answer hope this helps AJ

ankitesh gupta · Answer

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ankit singh · Answer

H 2 ​ ( g ) + S ( s ) ⇔ H 2 ​ S ( g ) 0.2 2 0 0.2-x 2-x x ∵ Δ n = 0 , K p ​ = K c ​ K c ​ = [ H 2 ​ ] [ H 2 ​ S ] ​ = ( 0 . 2 − x ) x ​ = 6 . 8 × 1 0 − 2 On solving this, we get = > x = 1 . 2 7 × 1 0 − 2 P H 2 ​ S ​ = [ x ] R T = 1 . 2 7 × 1 0 − 2 × 0 . 0 8 2 1 × ( 2 7 3 + 9 0 ) = 0 . 3 8 a t m

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At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 . If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 .

If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

3 Answers

H2(g)+S(s)⇔H2S(g)
0.2 2 0
0.2-x 2-x x

∵Δn=0,Kp=Kc

Kc=[H2][H2S]=(0.2−x)x=6.8×10−2

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At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 . If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 . If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

3 Answers

H2​(g)+S(s)⇔H2​S(g)0.2 2 00.2-x 2-x x ∵Δn=0,Kp​=Kc​ Kc​=[H2​][H2​S]​=(0.2−x)x​=6.8×10−2

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At 90°C , the following equilibrium is established : H2 (g) + S(s) = H2 S(g) Kp = 6.8 × 10 ^2 .

If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to 90°C in a 1.0 litre vessel, what will be the partial pressure of H2 S at equilibrium?

H2(g)+S(s)⇔H2S(g)
0.2 2 0
0.2-x 2-x x

∵Δn=0,Kp=Kc

Kc=[H2][H2S]=(0.2−x)x=6.8×10−2