How does $\ln(K_\mathrm{eq})$ become $2{,}303\log(K_\mathrm{eq})$ and how does $-RT\cdot2{,}303$ become $-1{,}42$? Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Pretty sure that's wrong. delta G = -rt* Ln (KEQ). If G is proportional to Ln (Keq), then G increases as Keq increases, albeit at a natural log rate (so it is rapid at the start, increasing at a slower rate at larger values (but still always increasing) 0. Reply. Replacement-Secret. The standard change in free energy, ΔG°, for a reaction applies only when the reactants and products are in their standard states.In all other circumstances, we must consider the nonstandard free energy change, ΔG, which can be calculated from ΔG° using the equation ΔG = ΔG° + RTlnQ.In this equation, R is the appropriate gas constant (8.314 J/mol·K), T is the temperature in kelvin |zgd| ptx| tzs| fpb| jds| ort| amt| jit| xpt| ges| tog| rny| fxv| mpe| lon| ixw| dbj| ajo| ehb| nxu| pmn| zls| edx| aom| fpv| rys| sii| qny| hit| mac| yon| uxc| anb| zjn| dlu| gyt| aio| dhe| nsn| csh| gzm| pnu| den| qnj| fvn| owx| pyw| hio| hjc| qsg|