Question

For the following reaction at a particular temperature, according to the equations
$2N_2{O}_5\to 4N{O}_2+O_2$
$2N{O}_2+\frac{1}{2}{O}_2\to N_2{O}_5$

• A. $E_1>E_2$
• B. $E_1<E_2$
• C. $E_1=2E_2$
• D. $\sqrt{E_1{E}_2^2}=1$

Answer 1

The correct option is A $E_1>E_2$

$\Delta H=E_1-E_b$
$-30=249-E_b$
$\therefore E_b=249+30=279kJmo{l}^{-1}$
The reaction
(i) $2N_2{O}_5\to 4N{O}_2+O_2$ is of $(C\to A+B)$ type
(ii) $2N{O}_2+\frac{1}{2}{O}_2\to N_2{O}_5$ is of $(A+B\to C)$
And for the reaction $(A+B\to C)$
$E_{a\left(f\right)}$ is 249 and $E_{a\left(b\right)}$ is 279 kJ $mo{l}^{-1}$
i.e., $E_1$ for reaction (i) > $E_2$ for reaction (ii).