For the reaction IO_3^- (aq) + 51^- (aq) + 6H^+(aq) rightarrow 3I_2(aq) + 3H_2O(

Question

For the reaction IO_3^- (aq) + 51^- (aq) + 6H^+(aq) rightarrow 3I_2(aq) + 3H_2O(l) the rate of disappearance of Gamma(aq) at a particular time and concentration is 2.6 Times 10^-3 mol/(L s). What is the rate of appearance of l_2(aq)? 4.3 Times 10^-3 mol/(L s) 7.8 Times 10^-3 mol/(L. s) -1.6 Times 10^-3 mol/(L s) 1.6 Times 10^-3 mol/(L s) 6.4 Times 10^-3 mol/(L s) For the first-order reaction 1./2 N_2O_4(g) rightarrow NO_2(g); Delta H = 28 kJ/mol the activation energy is 53 kJ/mol. What is the activation energy for the reverse reaction? -25 kJ/mol 81 kJ/mol -53 kJ/mol 25 kJ/mol 53 kJ/mol

Explanation / Answer

Answer – 3) We are given, reaction –

IO3-(aq) + 5I-(aq) + 6H+(aq) -----> 3I2(aq) + 3H2O(l)

Rate of disappearance of I-(aq) = 2.6*10-3 mol /(L.s)

So, - [ I-(aq)] /t = 2.6*10-3 mol /(L.s)

We know ,

Rate = - [ IO3-(aq)] /t = -1/5 [ I-(aq)] /t = -1/6 [ H+(aq)] /t

= 1/3[ I2(aq)] /t = 1/3[ H2O(l)] /t

So, -1/5 [ I-(aq)] /t = 1/3[ I2(aq)] /t

So, [ I2(aq)] /t = 3/5[ I-(aq)] /t

                            = 3/5*2.6*10-3 mol /(L.s)

Rate of appearance of I2 = 1.6*10-3 mol /(L.s)

Rate of disappearance is showing with negative since the concentration change is coming negative and rate is never negative. Rate of appearance is positive, since there is concentration change is coming positive.

4) We are given the reaction with H = 28 kJ/mol and activation energy is 53 kJ/mol, so the reverse reaction

So this is the endothermic reaction and the activation energy of the reverse reaction is less than the activation energy for the forward reaction.

So, activation energy of the reverse reaction = activation energy for the forward reaction - H

= 53 kJ/mol – 28 kJ/mol

= 25 kJ/mol

So answer is D) 25 kJ/mol

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