CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
10002.303 K
loader
C
1000 K
loader
D
20002.303 K
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image