A) the value of\[{{K}_{p}}\]increases with increase in temperature
B) the forward reaction gives out heat
C) there are more moelcules on the right hand side of the chemical equation than on the left
D) the reaction proceeds ten times faster at 450 K than 400 K
Correct Answer: B
Solution :
Variation of equilibrium constant with temperature can be express as: \[2.303\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{\Delta H}{R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}.{{T}_{2}}} \right]\] when \[{{K}_{1}}>{{K}_{2}}\]and \[{{T}_{1}}>{{T}_{2}}\] then \[\Delta H=-ve\]You need to login to perform this action.
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