A) \[\frac{{{K}_{p}}}{RT}={{({{K}_{c}})}^{\Delta {{n}_{g}}}}\]
B) \[\frac{{{K}_{c}}}{RT}={{({{K}_{p}})}^{\Delta {{n}_{g}}}}\]
C) \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]
D) \[{{K}_{c}}={{K}_{p}}{{(RT)}^{\Delta {{n}_{g}}}}\]
Correct Answer: C
Solution :
The relationship between\[{{K}_{c}}\]and\[{{K}_{p}}\]in gaseous equilibrium is\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\]You need to login to perform this action.
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