Q: The decomposition of phosphine \$(PH_3)\$ on tungsten at low pressure is a first-order reaction. It is because the:

The decomposition of \$PH_3\$ on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of \$PH_3\$. Hence, rate is proportional to the surface coverage.

Q: A first order reaction has a specific reaction rate of \$10^{-2} sec^{-1}\$. How much time will it take for \$20\ g\$ of the reactant to reduce to \$5\ g\$?

\$\displaystyle t = \dfrac {2.303}{k}log\left( \dfrac {a}{a-x} \right)\$\$\displaystyle t = \dfrac {2.303}{10^{-2} \: /s}log\left( \dfrac {20 \: g}{5 \: g}\right) \$\$\displaystyle t = 138.6\$ secHence, the time required will be 138.6 sec.

Question 1

Activation energy of a reaction:

Accepted Answer

Activation energy is defined as the minimum amount of extra energy required by a reacting molecule to get converted into products. Activation energy can't be altered by temperature. Catalysts are molecules that speed up reactions by decreasing the activation energy by providing an alternate path. The activation energy depends on the nature of the chemical transformation that takes place.

Question 2

The decomposition of phosphine $(PH_3)$ on tungsten at low pressure is a first-order reaction. It is because the:

Accepted Answer

The decomposition of $PH_3$ on tungsten at low pressure is a first order reaction because surface area coverage is proportional to&#10;partial pressure of $PH_3$. Hence, rate is proportional to the surface coverage.

Question 3

A first order reaction has a specific reaction rate of $10^{-2} sec^{-1}$. How much time will it take for $20\ g$ of the reactant to reduce to $5\ g$?

Accepted Answer

$\displaystyle  t = \dfrac {2.303}{k}log\left( \dfrac {a}{a-x} \right)$$\displaystyle  t = \dfrac {2.303}{10^{-2} \: /s}log\left( \dfrac {20 \: g}{5 \: g}\right) $$\displaystyle  t = 138.6$ secHence, the time required will be 138.6 sec.

Chemical Kinetics

Chemistry
4141 Views

Activation energy of a reaction:

The decomposition of phosphine $(P H_{3})$ on tungsten at low pressure is a first-order reaction. It is because the:

A first order reaction has a specific reaction rate of $1 0^{- 2} s e c^{- 1}$ . How much time will it take for $20 g$ of the reactant to reduce to $5 g$ ?

Chemical Kinetics

Chemistry 4141 Views

Activation energy of a reaction:

The decomposition of phosphine (PH3​) on tungsten at low pressure is a first-order reaction. It is because the:

A first order reaction has a specific reaction rate of 10−2sec−1. How much time will it take for 20 g of the reactant to reduce to 5 g?

Chemistry
4141 Views

The decomposition of phosphine $(P H_{3})$ on tungsten at low pressure is a first-order reaction. It is because the:

A first order reaction has a specific reaction rate of $1 0^{- 2} s e c^{- 1}$ . How much time will it take for $20 g$ of the reactant to reduce to $5 g$ ?