The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by
\(Rate = k [CH_3OCH_3]^{\frac 32}\)
The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
\(Rate = k(p_{CH_3OCH_3})^{\frac 32}\)
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants?
\(If\ pressure \ is \ measured \ in\ bar\ and\ time \ in\ minutes\, \ then\)
\(Unit \ of\ rate = bar \ min^{-1}\)
\(Rate = k(p_{CH_3OCH_3})^{\frac 32}\)
\(k = \frac {Rate}{(p_{CH_3OCH_3})^{\frac 32}}\)
\(Therefore, \ unit\ of \ rate\ constants = \frac {bar \ min^{-1}}{bar^{\frac 32}}\)
= \(bar^{-\frac 12} min^{-1}\)
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
For $\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftharpoons 2 \mathrm{AB}$ $\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Which of the following statement is correct?
The rate of a chemical reaction is defined as the change in concentration of any one of the reactants or products per unit time.
Consider the reaction A → B,
Rate of the reaction is given by,
Rate = −d[A]/ dt=+d[B]/ dt
Where, [A] → concentration of reactant A
[B] → concentration of product B
(-) A negative sign indicates a decrease in the concentration of A with time.
(+) A positive sign indicates an increase in the concentration of B with time.
There are certain factors that determine the rate of a reaction: