Question

# The rate constants of a reaction at $$500K$$ and $$700K$$ are $$0.02{s}^{-1}$$ and $$0.07{s}^{-1}$$ respectively. Calculate the values of $$Ea$$ and $$A$$:

Solution

## At $$500K$$ , $$K=0.02{ s }^{ -1 }$$At $$700K,\quad k=0.07{ s }^{ -1 }$$ Calculation of $${ E }_{ a }$$ and A can be done using Arrhenius equation $$k=A{ e }^{ { -E }_{ a }/RT }\\ \ln { k } =\ln { A } +\left( \cfrac { -{ E }_{ a } }{ RT } \right)$$ $$At\ 500\ K,\\ \ln { 0.02 } =\ln { A } +\cfrac { -{ E }_{ a } }{ 500R } \\ \ln { A } =\ln { (0.02) } +\cfrac { { E }_{ a } }{ 500R } \longrightarrow (1)\\ At\quad 700K\\ \ln { A } =\ln { (0.07) } +\cfrac { { E }_{ a } }{ 700R } \longrightarrow (2)$$ Equating (1) and (2)$$\ln { (0.02) } +\cfrac { { E }_{ a } }{ 500R } =\ln { (0.07)+\cfrac { { E }_{ a } }{ 700R } } \\ \cfrac { 1 }{ 100R } \left[ \cfrac { { E }_{ a } }{ 5 } -\cfrac { { E }_{ a } }{ 7 } \right] =\ln { \left[ \cfrac { 0.07 }{ 0.02 } \right] } \\ { E }_{ a }=\cfrac { 35\times 100\times 8.314\times 1.2528 }{ 2 } \\ { E }_{ a }=18227.6\quad J\\ { E }_{ a }=18.2\quad KJ$$ Substituting value of $${ E }_{ a }$$ in (1),$$\ln { A } =\ln { (0.02) } +\cfrac { 18227.6 }{ 500\times 8.314 } \\ =(-3.9120)+4.3848\\ \ln { A } =0.4728\\ \log { A } =1.0889\\ A=antilog\quad (1.0889)\\ A=12.27$$Thus $$E_a$$ is 18.2 KJ and A is 12.27Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More