Medium
Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?
- ABoth area $I$ and $II$ would increase
- BBoth area $I$ and $II$ would decrease
- CArea $I$ would increase and area $II$ would decrease
- DArea $I$ would decrease and area $II$ would increase
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Similar Questions
For the reaction $A_2 + B_2 \rightleftharpoons 2 AB$,the activation energies $(E_a)$ for the forward and backward reactions are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. If a catalyst lowers the $E_a$ for both reactions by $100 \ kJ \ mol^{-1}$,which of the following statements is correct?
According to the Arrhenius equation,the slope of the $\log k$ vs. $\frac{1}{T}$ plot is . . . . . . .
The graph obtained between $\ln k$ ($k=$ Rate constant) on $y$-axis and $1/T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$
Define the following terms:
$(a)$ Activation energy
$(b)$ Activated complex
$(a)$ Activation energy
$(b)$ Activated complex
Medium
View SolutionThe Arrhenius equation for the rate constant is $k = Ae^{-E_a/RT}$. $A$ chemical reaction will proceed more rapidly if there is a decrease in
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