For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

In a reaction, for every $10^{\circ} C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} C$ to $100^{\circ} C$, the rate of the reaction will become $:-$ (in $\times$)

For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

For the gaseous reactions:
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$
Calculate the approximate temperature at which $k_1 = k_2$ in $K$.

For a chemical reaction,if the temperature is increased from $25\,^{\circ}C$ to $55\,^{\circ}C$,the rate of reaction will change by a factor of (Assume $\mu = 3$).

$A \rightarrow B$
The rate constants of the above reaction at $200 \, K$ and $300 \, K$ are $0.03 \, min^{-1}$ and $0.05 \, min^{-1}$ respectively. The activation energy for the reaction is $.... \, J$ (Nearest integer).
(Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$,$\log 5 = 0.70$,$\log 3 = 0.48$,$\log 2 = 0.30$)