Saved Bookmarks
| 1. |
At `450 K, K_(p)=2.0xx10^(10)//` bar for the given reaction at equilibrium. `2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)` What is `K_(c )` at this temperature? |
|
Answer» Correct Answer - `7.47 xx 10^(11) M^(-1)` For the given reaction, `Deltan = 2-3 = -1` `T = 450 K` `R = 0.0831 "bar" L "bar" K^(-1) "mol"^(-1)` `K_(P) = 2.0 xx 10^(10) "bar"^(-1)` We know that, `K_(P) = K_(c) (RT) Deltan` `rArr 2.0 xx 10^(10) "bar"^(-1) = K_(c)(0.0831 L "bar"K^(-1) "mol"^(-1) xx 450 K)^(-1)` `rArr K_(c) = (2.0 xx 10^(10) "bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))` `= 2.0 xx 10^(10) "bar"^(-1) = K_(c) (0.0831L "bar" K^(-1) "mol"^(-1) xx 450K)` `rArr K_(c) = (2.0 xx 10^(10)"bar"^(-1))/((0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)^(-1))` `= (2.0 xx 10^(10) "bar"^(-1))(0.0831 L "bar" K^(-1) "mol"^(-1) xx 450 K)` `= 74.79 xx 10^(10) L "mol"^(-1)` `= 7.48 xx 10^(11) L "mol"^(-1)` `= 7.48 xx 10^(11) M^(-1)` |
|