The temperature in degrees Celsius that the reaction will go twice as fast is 32.4 ⁰C.
<h3>Temperature when the reaction rate is twice faster</h3>
ln(k₂/k₁) = E/R(1/T₁ - 1/T₂)
where;
- T₁ is initial temperature = 21 ⁰C = 294 K
- k₁ is initial rate
- k₂ is final rate
- T₂ is final temperature
- E is activation energy
- R ideal gas constant
when rate is twice faster, k₂ = 2k₁ = 2(0.011) s⁻¹ = 0.022 s⁻¹
ln(0.022/0.011) = (45,500/8.31)(1/294 - 1/T₂)
0.693 = 5475.33(1/294 - 1/T₂)
1.2657 x 10⁻⁴ = 1/294 - 1/T₂
1/T₂ = 1/294 - 1.2657 x 10⁻⁴
1/T₂ = 3.275 x 10⁻³
T₂ = 305.4 K = 32.4 ⁰C
Thus, the temperature in degrees Celsius that the reaction will go twice as fast is 32.4 ⁰C.
Learn more about rate of reaction here: brainly.com/question/24795637
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