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प्रश्न
An equilibrium constant of 3.2 × 10-6 for a reaction means, the equilibrium is
विकल्प
largely towards forward direction
largely towards reverse direction
never established
none of these
उत्तर
largely towards reverse direction
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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
The equilibrium constants of the following reactions are:
\[\ce{N2 + 3H2 <=> 2NH3}\]; K1
\[\ce{N2 + O2 <=> 2NO}\]; K2
\[\ce{H2 + 1/2O2 <=> H2O}\]; K3
The equilibrium constant (K) for the reaction;
\[\ce{2NH3 + 5/2 O2 <=> 2NO + 3H2O}\], will be
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.
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.
\[\ce{3H2S (g) <=> 3H2 (g) + 3/2 S2 (g)}\]
The equilibrium for the dissociation of XY2 is given as,
\[\ce{2 XY2 (g) <=> 2 XY (g) + Y2 (g)}\]
if the degree of dissociation x is so small compared to one. Show that 2 Kp = PX3 where P is the total pressure and Kp is the dissociation equilibrium constant of XY2.
The partial pressure of carbon dioxide in the reaction
\[\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\] is 1.017 × 10-3 atm at 500°C. Calculate Kp at 600°C for the reaction. H for the reaction is 181 KJ mol-1 and does not change in the given range of temperature.
The equilibrium constant Kp for the reaction \[\ce{N2 (g) + 3H2 (g) <=> 2NH3 (g)}\] is 8.19 × 102 at 298 K and 4.6 × 10-1 at 498 K. Calculate ∆H° for the reaction.
