Which statement is true with respect to a catalyst?

For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

The first order rate constant for a certain reaction increases from $1.667 \times 10^{-6} \ s^{-1}$ at $727 \ ^oC$ to $1.667 \times 10^{-4} \ s^{-1}$ at $1571 \ ^oC$. The rate constant at $1150 \ ^oC$,assuming constancy of activation energy over the given temperature range is [Given : $\log \ 19.9 = 1.299$ ]

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log k = -(2000) \frac{1}{T} + 6.0$. The pre-exponential factor $A$ and the activation energy $E_{a}$,respectively,are

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......$kJ/mol$.