If the temperature coefficient displaystyle left -frac-partial E-partial T- right- is zero a cell reaction then Delta S is-zeropositivenegativeindependent of displaystyle left -frac-partial E-partial T- right-

Question

Toppr Admin · Accepted Answer

-x394-S-x394-S-nF-dEdT-p-x394-S-x221D-dEdT-pSo when -dEdT-p-0- -x394-S-0-x394-H-x2212-nF-Ecell-x2212-T-dEdT-p- so -x394-H-x2260-0-x394-G-x394-H-x2212-T-x394-S- if -x394-S-0-xA0-xA0-x394-G-x394-H-x2260-0

If the temperature coefficient $(\frac{\partial E}{\partial T})$ is zero for a cell reaction then $Δ S$ is:
zero
positive
negative
independent of $(\frac{\partial E}{\partial T})$

$(Δ S)$
$Δ S = n F {(\frac{d E}{d T})}_{p}$
$Δ S \propto {(\frac{d E}{d T})}_{p}$
So when ${(\frac{d E}{d T})}_{p} = 0$ , $Δ S = 0$
$Δ H = - n F [E_{c e l l} - T {(\frac{d E}{d T})}_{p}]$ , so $Δ H \neq 0$
$Δ G = Δ H - T Δ S$ ; if $Δ S = 0$ , $Δ G = Δ H \neq 0$

If the temperature coefficient (∂E∂T) is zero for a cell reaction then ΔS is:zeropositivenegativeindependent of (∂E∂T)

(ΔS)ΔS=nF(dEdT)pΔS∝(dEdT)pSo when (dEdT)p=0, ΔS=0ΔH=−nF[Ecell−T(dEdT)p], so ΔH≠0ΔG=ΔH−TΔS; if ΔS=0, ΔG=ΔH≠0

If the temperature coefficient $(\frac{\partial E}{\partial T})$ is zero for a cell reaction then $Δ S$ is:
zero
positive
negative
independent of $(\frac{\partial E}{\partial T})$

$(Δ S)$
$Δ S = n F {(\frac{d E}{d T})}_{p}$
$Δ S \propto {(\frac{d E}{d T})}_{p}$
So when ${(\frac{d E}{d T})}_{p} = 0$ , $Δ S = 0$
$Δ H = - n F [E_{c e l l} - T {(\frac{d E}{d T})}_{p}]$ , so $Δ H \neq 0$
$Δ G = Δ H - T Δ S$ ; if $Δ S = 0$ , $Δ G = Δ H \neq 0$