The slope of the line for the graph of log k versus(1/T) for the reaction, N2O5→2NO2+(1/2)O2 is - 5000. Calculate the energy of activation of the reaction(kJ K-1mol-1)

Question

The slope of the line for the graph of log k versus(1/T) for the reaction, N2O5→2NO2+(1/2)O2 is - 5000. Calculate the energy of activation of the reaction(kJ K-1mol-1) (A) 95.7 (B) 9.57 (C) 957 (D) None

Tardigrade Admin · Accepted Answer

K=Ae-Ea/RT In K=In A-(Ea/RT) 2.303,log K=-(Ea/RT)+2.303 log A log K=-(Ea/2.303RT)+log A Slope (Ea/2.303R)=5000 Ea=5000×8.31×2.303 =95.7kJ k-1mol-1

Q. The slope of the line for the graph of log $k$ versus $\frac{1}{T}$ for the reaction, $N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2}$ is $- 5000.$ Calculate the energy of activation of the reaction $(k J K^{- 1} m o l^{- 1})$

$95.7$

$9.57$

$957$

None

$K = A e^{- E_{a / RT}}$
In $K = I n A - \frac{E _{a}}{RT}$
$2.303, l o g K = - \frac{E _{a}}{RT} + 2.303 l o g A$
$l o g K = - \frac{E _{a}}{2.303 RT} + l o g A$
Slope $\frac{E _{a}}{2.303 R} = 5000$
$E_{a} = 5000 \times 8.31 \times 2.303$
$= 95.7 k J k^{- 1} m o l^{- 1}$

Q. The slope of the line for the graph of log k versusT1​ for the reaction, N2​O5​→2NO2​+21​O2​ is −5000. Calculate the energy of activation of the reaction(kJK−1mol−1)

95.7

9.57

957