
An exothermic reaction is represented by the graph:
A.
B.
C.
D.
Answer
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Hint: The relation between equilibrium constant, ${K_p}$ and Temperature is given by Van’t Hoff’s equation. The equation relates the change in the equilibrium constant of a reaction with the change in temperature and the standard enthalpy change of a reaction. And for an exothermic reaction, the standard enthalpy change is always negative. This is because energy is released during an exothermic reaction.
Complete step by step answer:
The van’t Hoff equation was proposed by a Dutch chemist Jacobus Henricus van’t Hoff in the year 1884. The van’t Hoff’s equation is given by:
$\ln {K_p} = \dfrac{{\Delta {H^0}}}{{RT}} + \dfrac{{\Delta {S^0}}}{R}$, here $\Delta {H^0}$ is the change in standard enthalpy, R is the universal gas constant, T is the temperature, ${K_p}$ is the equilibrium constant, and $\Delta {S^0}$ is the change in the entropy.
If you look at the van’t Hoff equation it is in the form $y = mx + c$ where m is the slope of the reaction and c is the intercept.
So comparing both the equations the slope of the reaction is $ - \dfrac{{\Delta {H^0}}}{R}$ , since for an exothermic change is standard enthalpy is negative, it will give us a positive slope.
So, the correct answer is Option A.
Note: The equilibrium constant of a reaction is the value of the reaction quotient of that chemical reaction at a chemical equilibrium, which is a state approached by a dynamic chemical system after sufficient time has passed at which its composition does not change further. In simple words, an equilibrium is attained when the rate of the forward reaction is equal to the rate of backward reaction.
Complete step by step answer:
The van’t Hoff equation was proposed by a Dutch chemist Jacobus Henricus van’t Hoff in the year 1884. The van’t Hoff’s equation is given by:
$\ln {K_p} = \dfrac{{\Delta {H^0}}}{{RT}} + \dfrac{{\Delta {S^0}}}{R}$, here $\Delta {H^0}$ is the change in standard enthalpy, R is the universal gas constant, T is the temperature, ${K_p}$ is the equilibrium constant, and $\Delta {S^0}$ is the change in the entropy.
If you look at the van’t Hoff equation it is in the form $y = mx + c$ where m is the slope of the reaction and c is the intercept.
So comparing both the equations the slope of the reaction is $ - \dfrac{{\Delta {H^0}}}{R}$ , since for an exothermic change is standard enthalpy is negative, it will give us a positive slope.
So, the correct answer is Option A.
Note: The equilibrium constant of a reaction is the value of the reaction quotient of that chemical reaction at a chemical equilibrium, which is a state approached by a dynamic chemical system after sufficient time has passed at which its composition does not change further. In simple words, an equilibrium is attained when the rate of the forward reaction is equal to the rate of backward reaction.
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