Advertisements
Advertisements
प्रश्न
1 mol of CH4, 1 mole of CS2 and 2 mol of H2S are 2 mol of H2 are mixed in a 500 ml flask. The equilibrium constant for the reaction Kc = 4 x 10-2 mol2 lit-2. In which direction will the reaction proceed to reach equilibrium?
उत्तर
CH4(g) + 2 H2S (g) ⇌ CS2(g) + 4H2(g)
Kc = 4 x 10-2 mol2 lit-2
Volume = 500 ml = `1/2`L
[CH4]in = `(1 "mol")/(1/2 "L")` = 2 mol L-1
[CH2]in = `(1 "mol")/(1/2 "L")` = 2 mol L-1
`["H"_2"S"]_"in" = (2 "mol")/(1/2 "L")`= 4 mol L-1
`["H"_2]_"in" = (2 "mol")/(1/2 "L")`= 4 mol L-1
Q = `(["CS"_2]["H"_2]^4)/(["CH"_4]["H"_2"S"]^2)`
`= ([2][4]^4)/([2][2]^2)`
= 64 mol2 lit-2
Q > KC.
∴ The reaction will proceed in the reverse direction to reach the equilibrium.
APPEARS IN
संबंधित प्रश्न
K1 and K2 are the equilibrium constants for the reactions respectively.
\[\ce{N2(g) + O2(g) <=>[K1] 2NO(g)}\]
\[\ce{NO(g) + O2(g) <=>[K2] 2NO2(g)}\]
What is the equilibrium constant for the reaction \[\ce{NO2(g) <=> 1/2 N2(g) + O2(g)}\]
In the equilibrium,
\[\ce{2A(g) <=> 2B(g) + C2(g)}\]
the equilibrium concentrations of A, B and C2 at 400 K are 1 × 10–4 M, 2.0 × 10–3 M, 1.5 × 10–4 M respectively. The value of KC for the equilibrium at 400 K is
In which of the following equilibrium, Kp and Kc are not equal?
For a given reaction at a particular temperature, the equilibrium constant has a constant value. Is the value of Q also constant? Explain.
Derive the relation between Kp and Kc.
One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, Calculate the value of equilibrium constant.
For the reaction
\[\ce{SrCO3(s) <=> SrO(s) + CO2(g)}\]
the value of equilibrium constant Kp = 2.2 × 10-4 at 1002 K. Calculate Kc for the reaction.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate Kc and Kp.
At particular temperature Kc = 4 × 10-2 for the reaction, \[\ce{H2S (g) <=> H2(g) +1/2 S2(g)}\]. Calculate the Kc for the following reaction.
\[\ce{2H2S (g) <=> 2H2 (g) + S2 (g)}\]
At particular temperature Kc = 4 × 10-2 for the reaction, \[\ce{H2S (g) <=> H2(g) +1/2 S2(g)}\]. Calculate the Kc for the following reaction.
\[\ce{3H2S (g) <=> 3H2 (g) + 3/2 S2 (g)}\]
