Question

For the reaction:

$\left[Cr\right(H_{2}O{)}_6{]}^{3+}+\left[SC{N}^{\ominus }\right]\to [Cr9H_{2}O{)}_{5}NCS{]}^{2+}{H}_{2}O$

The rate law is : $r=k\left[Cr\right(H_{2}O{)}_6^{3+}\left[SC{N}^{\ominus }\right].$

The value of k is $2.0\times {10}^{-6}Lmo{l}^{-1}{s}^{-1}$ at 14${}^0$C and $2.2\times {10}^{-5}Lmo{l}^{-1}{s}^{-1}at{30}^{\circ }C.$ What is the value of $E_a$?

• A. 26 kcal $mo{l}^{-1}$
• B. 2.6 kcal $mo{l}^{-1}$
• C. 2600 kcal $mo{l}^{-1}$
• D. 260 kcal $mo{l}^{-1}$

Answer 1

The correct option is A 26 kcal $mo{l}^{-1}$

Using Arrhenius equation $T_1$ = 273 + 14 = 287 K, $T_2$ = 273 + 30 = 303 K

$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1\times T_2}\right]$

$log\frac{2.2\times {10}^{-5}}{2.0\times {10}^{-6}}=\frac{E_a}{2.303\times 2\times {10}^{-3}}\left(\frac{16}{287\times 303}\right)$

$E_a=\frac{log\left(11\right)\times 2.303\times 2\times {10}^{-3}\times 287303}{16}$

$=\frac{1.0414\times 2.303\times 2\times 287\times 303}{16000}$

$=26kcalmo{l}^{-1}$