NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NTA Abhyas 2020 · Chemistry · Chemical Equilibrium
What is the equilibrium expression for the reaction P_ 4 (s) + 5 O_...
What is the equilibrium expression for the reaction
{\text{P}_{4}\left(\text{s}\right) + 5{\text{O}_{2} ( \text{g} ) ⇌ {\text{P}_{4}{\text{O}_{10} ( \text{s} )?
{\text{P}_{4}\left(\text{s}\right) + 5{\text{O}_{2} ( \text{g} ) ⇌ {\text{P}_{4}{\text{O}_{10} ( \text{s} )?
Options
- A{K}_{c} = \dfrac{\left[{P}_{4} {O}_{10}\right]}{\left[{P}_{4}\right]{\left[{O}_{5}\right]}^{5}}
- B{\text{K}_{\text{c}} = \dfrac{[ {\text{P}_{4}{\text{O}_{10} ]}{5\left[{\text{P}_{4}\right] [ {\text{O}_{2} ]}
- C{\text{K}_{\text{c}} = {\left[{\text{O}_{2}\right]}^{5}
- D{\text{K}_{\text{c}} = \dfrac{1}{{\left[{\text{O}_{2}\right]}^{5}}
Correct answer
D. {\text{K}_{\text{c}} = \dfrac{1}{{\left[{\text{O}_{2}\right]}^{5}}
Step-by-step solution
In the expression for equilibrium constant ( {\text{K}_{\text{p}}\text{o}\text{r}{\text{K}_{\text{c}} ) species in solid state are not written (., their molar concentrations are taken as 1)
{\text{P}_{4}\left(\text{s}\right) + 5{\text{O}_{2} ( \text{g} ) ⇌ {\text{P}_{4}{\text{O}_{10} ( \text{s} )
Thus, {\text{K}_{\text{c}} = \dfrac{1}{{\left[{\text{O}_{2}\right]}^{5}}
{\text{P}_{4}\left(\text{s}\right) + 5{\text{O}_{2} ( \text{g} ) ⇌ {\text{P}_{4}{\text{O}_{10} ( \text{s} )
Thus, {\text{K}_{\text{c}} = \dfrac{1}{{\left[{\text{O}_{2}\right]}^{5}}