Question

Rate of disappearance of reactant A at two different temperatures is given by $$A\rightleftharpoons B$$.
${\displaystyle \frac{-d\left[A\right]}{dt}}=\left(2\times {10}^{-2}{S}^{-1}\right)\left[A\right]-\left(4\times {10}^{-3}{S}^{-1}\right)\left[B\right];300K$
${\displaystyle \frac{-d\left[A\right]}{dt}}=\left(4\times {10}^{-2}{S}^1\right)\left[A\right]-\left(16\times {10}^{-4}{S}^{-1}\right)\left[B\right];400K$
 Calculate the heat of reaction in the given temperature range, when equilibrium is set up.
(A) 8.03 kJ
(B) 16.06 kJ
(C) 32.12 kJ
(D) None of these

Answer 1

Hint:This question is solved by using the equation of equilibrium constant. As in this question we have given the rate of disappearance of reactant A at different temperatures, we have to find equilibrium constant for both the temperature and then we can find the heat of reaction. Then by using the Arrhenius equation we can find the heat of reaction.

Complete solution: First, we have to find the value of equilibrium constant for given different temperatures. Equilibrium constant is the value of reaction quotient at chemical equilibrium.
$$A\rightleftharpoons B$$
For 300K temperature; $K_1={\displaystyle \frac{2\times {10}^{-2}{S}^{-1}}{4\times {10}^{-3}{S}^{-1}}}$ = 5
For 400k temperature; $K_2={\displaystyle \frac{4\times {10}^{-2}{S}^{-1}}{16\times {10}^{-4}{S}^{-1}}}$ = 25
Here, above we got the values of the equilibrium constants at different temperatures. It is clear from the above value K (Equilibrium constant) that the temperature of the reaction. Now, by using the Arrhenius equation we can calculate the heat of the reaction at a given temperature range.
$\log {\displaystyle \frac{K_2}{K_1}}={\displaystyle \frac{\mathrm{\Delta }H}{2.303\times R}}\left[{\displaystyle \frac{1}{T_1}}-{\displaystyle \frac{1}{T_2}}\right]$
Where, K1 and K2 are the equilibrium constant at temperature 300 K (T1) and temperature 400 K (T2) respectively.
R is the gas constant.
Hence, putting the values in above equation we get,
$\log {\displaystyle \frac{25}{5}}={\displaystyle \frac{\mathrm{\Delta }H}{2.303\times 8.314J{K}^{-1}mo{l}^{-1}}}\left[{\displaystyle \frac{1}{300K}}-{\displaystyle \frac{1}{400K}}\right]$
Now, evaluating and rearranging the above equation we get the value of delta H.
$\log 5={\displaystyle \frac{\mathrm{\Delta }H\left(0.0008K^{-1}\right)}{19.1471J{K}^{-1}mo{l}^{-1}}}$
$\mathrm{\Delta }H=16.06KJ$
Hence, the heat of reaction for the given temperature range for disappearance of the reactant A would be
$\mathrm{\Delta }H=16.06KJ$.

Thus,Option B is correct

Note:Now it is clear from the above that as we can change the temperature for the disappearance of the reactant A the heat of reaction is depend on the two different equilibrium constants for the particular reaction. Hence, we can get the heat of reaction is $\mathrm{\Delta }H=16.06KJ$. Heat of reaction is the change in the enthalpy of the system during the reaction.