The following reaction is performed at $298\;K$

Question

$\begin{array}{1 1} R(298) ln(1.6 \times 10^{12} ) -86600 \\ 86600+R(298) ln(1.6 \times 10^{12}) \\ 86600 - \frac{ln(1.6 \times 10^{12})}{R(298)} \\ 0.5[2 \times 86,600 -R(298 ) ln(1.6 \times 10^{12})] \end{array} $