MCQ
For any spontaneous reaction, the equilibrium constant, $E _{\text {cell }}^{\circ}$ and $\Delta G$ respectively will be :
- A$<1,+$ ve, - ve
- ✓$>1,+v e,-v e$
- C$=1,+ ve ,+ ve$
- D$=1,- ve ,- ve$
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$A \to B$ ${k_1} = {10^{15}}{e^{ - 25000/8.34\,T}}$
$C \to D$ ${k_2} = {10^{14}}{e^{ - 15000/8.34\,T}}$
Calculate the approximate temperature at which $k_1 = k_2$ ...... $K$
| column $(I)$ | column $(II)$ | ||
| $(A)$ | Kohlraush law | $(i)$ | $\Lambda _{eq}^o = \Lambda _c^o + \Lambda _a^o$ |
| $(B)$ | Molar Conductivity |
$(ii)$ | $\Lambda _m = \frac{{K \times 1000}}{M}$ |
| $(C)$ | Degree of Dissociation |
$(iii)$ | $\alpha = {\Lambda _m}/\Lambda _m^o$ |
| $(D)$ | Dissociation Constant |
$(iv)$ | ${k_a} = C{\alpha ^2}/1 - \alpha $ |