Question

The table below gives the values for the rate constant, k, of the reaction between potassium hydroxide and bromoethane in ethanol at a series of temperatures. Use these data to determine the activation energy of the reaction.

Explain your answer in brief.

T/K 305.0 313.0 323.1 332.7 343.6 353.0

k/M-1 s-1 0.182 0.466 1.35 3.31 10.2 22.6

A 80J

B 90J

C 80 J mol-1

D 80 kJ mol-1

E 90 kJ mol-1

Answer #1

Calculate the values of ln K and 1/T

K | ln K | T | 1/T |

0.182 | -1.7037 | 305 | 0.00327 |

0.466 | -0.7636 | 313 | 0.00319 |

1.35 | 0.3 | 323.1 | 0.00309 |

3.31 | 1.197 | 332.7 | 0.00301 |

10.2 | 2.322 | 343.6 | 0.00291 |

22.6 | 3.118 | 353 | 0.00283 |

From the Arrhenius equation

ln K = (-E/R) *(1/T) + ln A

?Y = m*X + c

Slope = -(y2 - y1) / (x1 - x2)

= -(2.322 - 1.197)/(0.00301 - 0.00291)

= -11250 K-1

?Slope = - E/R

-11250 K-1 = - E/(8.314 J/mol·K)

?E = 93532.5 J/mol x 1kJ/1000J

?E = 93.5 kJ

?Option E is the correct answer (nearest)

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constant k and the temperature T in kelvins and
is typically written as
k=Ae−Ea/RT
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a constant called the frequency factor, and Eais
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However, a more practical form of this equation is
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Arrhenius equation shows the relationship between the rate constant
k and the temperature Tin kelvins and is
typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), Ais
a constant called the frequency factor, and Ea is
the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathmatically equivalent to
lnk1k2=EaR(1T2−1T1)
where k1 and k2 are the rate constants for a
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k and the temperature T in kelvins
and is typically written as
k=Ae−Ea/RT
where R is the gas constant (8.314 J/mol⋅K), A
is a constant called the frequency factor, and Ea
is the activation energy for the reaction.
However, a more practical form of this equation is
lnk2k1=EaR(1T1−1T2)
which is mathematically equivalent to
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