Question

# The rate constants $${ k }_{ 1 }$$ and $${ k }_{ 2 }$$ for two different reactions are $${ 10 }^{ 16 }{ e }^{ { -2000 }/{ T } }$$ and $${ 10 }^{ 15 }{ e }^{ { -1000 }/{ T } }$$ respectively. The temperature at which $${ k }_{ 1 }={ k }_{ 2 }$$ is:

A
2000 K
B
10002.303 K
C
1000 K
D
20002.303 K

Solution

## The correct option is B $$\dfrac { 1000 }{ 2.303 }$$ $$K$$$${ 10 }^{ 15 }{ e }^{ { -1000 }/{ T } }={ 10 }^{ 16 }{ e }^{ { -2000 }/{ T } }$$$$\dfrac { { e }^{ { -2000 }/{ T } } }{ { e }^{ { -1000 }/{ T } } } =\dfrac { { 10 }^{ 15 } }{ { 10 }^{ 16 } }$$$${ e }^{ { -1000 }/{ T } }={ 10 }^{ -1 }$$$$\log _{ e }{ { e }^{ { -1000 }/{ T } } } =\log _{ e }{ { 10 }^{ -1 } }$$$$\dfrac { -1000 }{ T } =2.303\log _{ 10 }{ { 10 }^{ -1 } } =-2.303$$$$T=\dfrac { 1000 }{ 2.303 } K$$Chemistry

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