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For an exothermic reaction, what happens...

For an exothermic reaction, what happens to the equilibrium constant if temperature is increased?

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To determine what happens to the equilibrium constant of an exothermic reaction when the temperature is increased, we can follow these steps: ### Step 1: Understand Exothermic Reactions Exothermic reactions are those that release heat. In a chemical equation, they can be represented as: \[ \text{Reactants} \rightleftharpoons \text{Products} + \text{Heat} \] ### Step 2: Apply Le Chatelier's Principle Le Chatelier's Principle states that if an external change is applied to a system at equilibrium, the system will adjust itself to counteract that change and restore a new equilibrium. ...
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Knowledge Check

  • For a given exothermic reaction , K_P and K_p' are the equilibrium constant at temperature T_1 and T_2 respectively .Assuming that heat of reaction is constant in temperature range between T_1 and T_2 , it is readly observed that

    A
    ` K_p gt K_p `
    B
    ` K_p lt K_p `
    C
    ` K_p = K_p `
    D
    ` K_p = (1)/(K_p')`

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    Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction DeltaG^(@)=-2.30RTlogk DeltaG^(@): Standing free energy change DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) …(ii) DeltaH^(@) : Standard heat of the reaction gt From eqns.(i) and(ii) -2RTlogk=DeltaH^(@)=TDeltaS^(@) DeltaS^(@) : standard entropy change implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R) Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope =(-DeltaH^(@))/(2.3R) amd y intercept =(DeltaS^(@))/(2.3R) If at temperature T_(1) equilibrium constant be k_(1) and at temperature T_(2) equilibrium constant be k_(2) then : implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R) ..(iv) implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R) ...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction For exothermic reaction if DeltaS^(@)lt0 then the sketch of log k vs (1)/(T) may be

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