The equilibrium constant `K_(c)` for the following reaction at `842^(@)`C is `7.90xx10^(-3)`. What is `K_(p)`at same temperature ? ` (1)/(2)F_(2(g)) h

Question

The equilibrium constant `K_(c)` for the following reaction at `842^(@)`C is `7.90xx10^(-3)`. What is `K_(p)`at same temperature ? ` (1)/(2)F_(2(g)) h

The equilibrium constant `K_(c)` for the following reaction at `842^(@)`C is `7.90xx10^(-3)`. What is `K_(p)`at same temperature ?
` (1)/(2)F_(2(g)) hArr F_(g)`
A. `8.64xx10^(5)`
B. `8.26xx10^(-4)`
C. `7.90xx10^(-2)`
D. `7.56xx10^(-2)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`K_(P)=(7.9xx10^(-3))xx(0.0821xx115)^(0.5)`
`K_(P)=7.56xx10^(-2)`

4 views Answered 2023-02-05 15:29