Which of the following will favour the reverse reaction in a chemical equilibrium reaction?
A. Increasing the concentration of one of the reactants
B. Increasing the concentration of one or more of the products
C. Removal of at least one of the products at regular interval
D. None of these
Answer: Option B
A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these

Join The Discussion