Rate of a reaction decreases with the passage of time as the concentration of reactants decrease. Conversely, rates generally increase when reactant concentrations increase. So, rate of a reaction depends upon the concentration of reactants.
Consider a general reaction
aA + bB → cC + dD
where a, b, c and d are the stoichiometric coefficients of reactants and products.
The rate expression for this reaction is
Rate ∝ [A]x [B]y
where exponents x and y may or may not be equal to the stoichiometric coefficients (a and b) of the reactants.
Above equation can also be written as Rate = k [A]x [B]y
-\(\frac{dR}{dT}\) = k [A]x [B]y
This form of equation is known as differential rate equation, where k is a proportionality constant called rate constant.
The rate of a gaseous reaction is given by the expression k [A][B]. If volume of the reaction vessel is suddenly increased to double the initial volume, what will be the reaction rate w.r.t to the original rate?
Answer & explanation
Correct answer: option 2
r = k[A][B]
If volume of the vessel is doubled then the concentration of the reactant gets halved.
r' = k[\(\frac{A}{2}\)][\(\frac{B}{2}\)]
r' = \(\frac{k}{4}\)[A][B]
r' = \(\frac{r}{4}\)