For the reversible reaction A ⇋ 2B, if the equilibrium constant K is 0.05 mole/litre; starting from initially 2 moles of A and zero moles of B, how many moles will be formed at equilibrium?

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. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$

A. 1.5 K B. 2.5 K C. $$\frac{{0.693}}{{\text{K}}}$$ D. 6.93 K

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