We know \(\log K=\log A-\frac{E_{a}}{2.303 R T}\)
\(\Rightarrow \quad \frac{{E}_{{a}}}{2.303 {RT}}=2.47 \times 10^{3}\)
\({E}_{{a}}=2.47 \times 10^{3} \times 2.303 \times \frac{8.314}{1000} \,{KJ} / {mole}\)
\(=47.29=47({Nearest} \text { integer })\)
(નજીકના પૂર્ણાંકમાં રાઉન્ડ ઑફ) $[$ ઉપયોગ કરો : $\left. R =8.31 \,J \,K ^{-1} \,mol ^{-1}\right]$
( $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$ આપેલ છે.)