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The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k1 and k2 respectively. The energy of activation for the two reactions are 152.30 kJ mol−1 and 157.7 kJ mol−1 as well as frequency factors are 1013 and 1014 respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be the same for the two reactions.
  • a)
    256 K
  • b)
    354 K
  • c)
    282 K
  • d)
    674 K
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The rate of decomposition for methyl nitrite and ethyl nitrite can be...
For methyl nitrite k1 = 1013 e[−152300/(8.314×T)]
For ethyl nitrite k2 = 1014 e[−157700/(8.314×T)]
If k1 = k2 then
1013 e[−152300/(8.314×T)] = 1014 e[−157700/(8.314×T)]
T = 282 K
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Community Answer
The rate of decomposition for methyl nitrite and ethyl nitrite can be...
Given data:
- Energy of activation for methyl nitrite decomposition (Ea1) = 152.30 kJ mol−1
- Energy of activation for ethyl nitrite decomposition (Ea2) = 157.7 kJ mol−1
- Frequency factor for methyl nitrite decomposition (A1) = 1013
- Frequency factor for ethyl nitrite decomposition (A2) = 1014

Calculating the temperature for equal rate constants:
- The Arrhenius equation is given as: k = A * exp(-Ea/RT)
- For the rate constants to be equal, k1 = k2
- Therefore, A1 * exp(-Ea1/RT) = A2 * exp(-Ea2/RT)
- Taking natural logarithm on both sides, we get: -Ea1/RT + ln(A1) = -Ea2/RT + ln(A2)
- Rearranging the equation, we get: (Ea2 - Ea1) / R = 1/T * (ln(A1) - ln(A2))
- Substituting the values, we get: (157.7 - 152.30) / 8.314 = 1/T * (ln(1013) - ln(1014))
- Solving for T, we get T = 282 K
Therefore, the temperature at which the rate constant will be the same for the two reactions is 282 K, which corresponds to option 'C'.
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The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k1 and k2 respectively. The energy of activation for the two reactions are 152.30 kJ mol−1 and 157.7 kJ mol−1 as well as frequency factors are 1013 and 1014 respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be the same for the two reactions.a)256 Kb)354 Kc)282 Kd)674 KCorrect answer is option 'C'. Can you explain this answer?
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