The reaction between ethyl iodide and hydroxide ion in ethanol(C2H5OH) solution
ID: 75545 • Letter: T
Question
The reaction between ethyl iodide and hydroxide ion in ethanol(C2H5OH) solution C2H5I+OH=C2H5OH+I has an activation energy of 86.8KJ/mol and a frequency factorof 2.10*10^11 M^(-1)S^(-1) Q: Equal volumes of a 0.0239 M KOH in ethanol solution and a0.0373 M C2H5I in ethanol solution are mixed. Assuming the reactionis first order in each reactant, what is the initial rate ofreaction at 35 degree celsius? Hint: Do your dilutions first! thank you so much! life saver rewarded if answer's satisfactory! The reaction between ethyl iodide and hydroxide ion in ethanol(C2H5OH) solution C2H5I+OH=C2H5OH+I has an activation energy of 86.8KJ/mol and a frequency factorof 2.10*10^11 M^(-1)S^(-1) Q: Equal volumes of a 0.0239 M KOH in ethanol solution and a0.0373 M C2H5I in ethanol solution are mixed. Assuming the reactionis first order in each reactant, what is the initial rate ofreaction at 35 degree celsius? Hint: Do your dilutions first! thank you so much! life saver rewarded if answer's satisfactory!Explanation / Answer
Formula: k = A e-Ea / RT Where k is the rate constant Ais the frequency factor Ea is the activation energy R isthe gas constant ( 8.3145 J / mol.K) T isthe kelvin temperature It is also mentioned that the reaction is first order ineach reactant Then Initial rate = k[OH-][C2H5I] Data: [C2H5I]= 0.0373 M [OH-] = 0.0239 M A = 2.10*10^11 M^(-1)S^(-1) Ea = 86.8KJ/mol T = 35 + 273 K =308 K From this data k value can be found at 350C. k = 2.10*10^11 M^(-1)S^(-1) e-86.8 * 1000 J/mol / 8.3145 J / mol.K * 308 K =2.10*10^11 M^(-1)S^(-1) * 1.90 * 10-15 = 3.99 * 10-4 M /s Now the k value is substituted in the expression, Initial rate = 3.99 *10-4 M /s ( 0.0373)( 0.0239 ) =3.56 * 10-7Related Questions
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