JEE MAIN - Chemistry (2005 - No. 33)
For a spontaneous reaction the ∆G , equilibrium constant (K) and $$E_{cell}^o$$ will be respectively
-ve, >1, +ve
+ve, >1, -ve
-ve, <1, -ve
-ve, >1, -ve
Explanation
NOTE : For spontaneous reaction $$\Delta G$$ should be negative. Equilibrium constant should be more than one
$$\left( {\Delta G = - 2.303\,RT\,\log {K_c},\,\,} \right.$$
If $${K_c} = 1\,\,$$ then $$\,\,\,\Delta G = 0;\,\,$$
If $${K_c} < 1$$ then $$\left. {\Delta G = + ve} \right).\,\,$$
Again$$\,\,\,\Delta G = - nFE_{cell}^ \circ $$
$$E_{cell}^ \circ \,\,\,$$ must be $$+ve$$ to have $$\Delta G - ve.$$
$$\left( {\Delta G = - 2.303\,RT\,\log {K_c},\,\,} \right.$$
If $${K_c} = 1\,\,$$ then $$\,\,\,\Delta G = 0;\,\,$$
If $${K_c} < 1$$ then $$\left. {\Delta G = + ve} \right).\,\,$$
Again$$\,\,\,\Delta G = - nFE_{cell}^ \circ $$
$$E_{cell}^ \circ \,\,\,$$ must be $$+ve$$ to have $$\Delta G - ve.$$
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