Question

Listed in the table are forward and reverse rate constants for the reaction
$2\,{\text{NO(g)}}\,\, \rightleftarrows {{\text{N}}_2}{\text{(g)}}\,\,{\text{ + }}\,{{\text{O}}_2}{\text{(g)}}\,$

Temperature (K)	${{\text{K}}_{\text{f}}}$ $\left( {{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}} \right)$	${{\text{K}}_{\text{r}}}$ $\left( {{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}} \right)$
$1400$	$0.29$	$1.1 \times {10^{ - 6}}$
$1500$	$1.3$	$1.4 \times {10^{ - 5}}$

Is the reaction endothermic or exothermic?
A. Exothermic
B. Endothermic
C. Can't predict
D. None of these

Answer 1

We can use the Le-chatlier principle. The process which absorbs heat during the reaction and decreases the temperature of the surrounding is known as endothermic process. The process which releases heat during the reaction and increases the temperature of the surrounding is known as exothermic process.

Formula used: ${\text{K = }}\,\dfrac{{{{\text{K}}_{\text{f}}}}}{{{{\text{K}}_{\text{r}}}}}$

Complete answer
According to the Le-chatlier principle if a reaction equilibrium is disturbed by changing the variable such as temperature, pressure number of mole, the reaction goes in that direction where the effect of change can be cancelled out.

Effect of temperature: Endothermic reactions require heat to occur so, on increasing temperature equilibrium shifts in forward direction and thus the value of equilibrium constant increases. Exothermic reactions release heat so, on increasing temperature equilibrium shifts in backward direction so, the value of equilibrium constant decreases.
The given temperature and rate constants are as follows:

Temperature (K)	${{\text{K}}_{\text{f}}}$ $\left( {{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}} \right)$	${{\text{K}}_{\text{r}}}$ $\left( {{{\text{M}}^{ - 1}}{{\text{s}}^{ - 1}}} \right)$
$1400$	$0.29$	$1.1 \times {10^{ - 6}}$
$1500$	$1.3$	$1.4 \times {10^{ - 5}}$

The relation between rate constants of forward and reverse direction is as follows:
${\text{K = }}\,\dfrac{{{{\text{K}}_{\text{f}}}}}{{{{\text{K}}_{\text{r}}}}}$
The equilibrium constant at $1400$K is as follows:
Substitute $0.29$ for ${{\text{K}}_{\text{f}}}$ and $1.1 \times {10^{ - 6}}$ for ${{\text{K}}_{\text{r}}}$.
$\Rightarrow {{\text{K}}_{1400}}{\text{ = }}\,\dfrac{{{\text{0}}{\text{.29}}}}{{1.1 \times {{10}^{ - 6}}}}$
$\Rightarrow {{\text{K}}_{1400}}{\text{ = }}2.6 \times {10^5}$
The equilibrium constant at $1500$K is as follows:
Substitute $1.3$ for ${{\text{K}}_{\text{f}}}$ and $1.4 \times {10^{ - 5}}$ for ${{\text{K}}_{\text{r}}}$.
$\Rightarrow {{\text{K}}_{1500}}{\text{ = }}\,\dfrac{{{\text{1}}{\text{.3}}}}{{1.4 \times {{10}^{ - 5}}}}$
$\Rightarrow {{\text{K}}_{1500}}{\text{ = }}9.3 \times {10^4}$
The equilibrium constant at $1400$ temperature is $2.6 \times {10^5}$ whereas on increasing temperature from $1400$ to $1500$, equilibrium constant is decreasing $9.3 \times {10^4}$. So, on increasing temperature equilibrium constant decreasing so, the reaction is exothermic.

Therefore, option (A) exothermic, is the correct answer.

Note: in endothermic reactions, on increasing temperature the rate of forward direction increases and thus rate of reverse direction decreases so, equilibrium shifts in forward direction. In exothermic reactions, on increasing temperature the rate of forward direction decreases and thus rate of reverse direction increases so equilibrium shifts in backward direction. The high temperature favors the endothermic reaction and low-temperature favour the exothermic reaction.