Advertisements
Advertisements
प्रश्न
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?
उत्तर
When Q > KC the reaction will proceed in the reverse direction, i.e, formation of reactants.
APPEARS IN
संबंधित प्रश्न
Which one of the following is incorrect statement ?
The values of Kp1 and Kp2; for the reactions,
X ⇌ Y + Z,
A ⇌ 2B are in the ratio 9 : 1 if degree of dissociation of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 10-2, and the equilibrium constant is 50. The rate constant for the reverse reaction is,
For the formation of Two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 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
Derive a general expression for the equilibrium constant Kp and Kc for the reaction, \[\ce{3H2(g) + N2(g) <=> 2NH3(g)}\].
Write the balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression.
`"K"_"C" = (["NH"_3]^4["O"_2]^5)/(["NO"]^4["H"_2"O"]^6)`
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.
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.
\[\ce{3H2S (g) <=> 3H2 (g) + 3/2 S2 (g)}\]