Solve this-Q33- Effect of increasing temperature on equilibrium constant is given by log K2 nbsp- ndash- log K1 nbsp- nbsp- -8710-H2-303R -160-1T2-1T1- Then for an endothermic nbsp-reaction the false statement is - nbsp- nbsp-1- nbsp- -160-1T2-1T1 nbsp- positive nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- -2- log K2 gt- log K1- nbsp- nbsp- nbsp- nbsp-3- nbsp- -8710-H- positive nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp- nbsp-4- K2 gt- K1- nbsp-

Dear Student, log #160;K2 log #160;K1 #160;= #160; #8710;H2.303R[1T2 1T1]In #160;endothermic #160;reaction, #160; #8710;H #160;is #160;+ ...

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Standard XII

Chemistry

Activation Energy

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Q33. Effect of increasing temperature on equilibrium constant is given by log K₂ – log K₁ = $\frac{∆ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ . Then for an endothermic reaction the false statement is :-

(1) $[\frac{1}{T_{2}} - \frac{1}{T_{1}}]$ = positive (2) log K₂ > log K₁

(3) $∆ H$ = positive (4) K₂ > K₁

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Solution

Dear Student,

$\log K_{2} - \log K_{1} = \frac{- ∆ H}{2.303 R} [\frac{1}{T_{2}} - \frac{1}{T_{1}}] In endothermic reaction, ∆ H is + ve On increasing temperature, endothermic reaction is favoured, so K_{2} > K_{1} . Also, \log K_{2} > \log K_{1} Here temperature is being increased, so T_{2} > T_{1} So, [\frac{1}{T_{2}} - \frac{1}{T_{1}}] = [\frac{T_{1} - T_{2}}{T_{1} T_{2}}] = - ve Hence, false statement is (1)$

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Solve this: ​Q33. Effect of increasing temperature on equilibrium constant is given by log K2 – log K1 = ∆H2.303R 1T2-1T1. Then for an endothermic reaction the false statement is :- (1) 1T2-1T1 = positive (2) log K2 > log K1 (3) ∆H= positive (4) K2 > K1

Dear Student, log K2-log K1 = -∆H2.303R[1T2-1T1]In endothermic reaction, ∆H is +veOn increasing temperature, endothermic reaction is favoured, so K2 > K1. Also, log K2 > log K1Here temperature is being increased, so T2 > T1So, [1T2-1T1] =[T1-T2T1T2] =-veHence, false statement is (1)