JEE Advance - Chemistry (2020 - Paper 2 Offline - No. 17)
Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.
At $$298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$$ mol-1,
$$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$
Assume that, the enthalpies and the entropies are temperature independent.
At $$298K:{\Delta _f}H^\circ [Sn{O_2}(s)] = - 581.0$$ mol-1,
$$\eqalign{ & {\Delta _f}H^\circ [(C{O_2})(g)] = - 394.0\,kJ\,mol{ ^{-1}} \cr & S^\circ [Sn{O_2}(s)] = 56.0J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [Sn(s)] = 52.0\,J\,K{ ^{-1}}mo{l^{ - 1}} \cr & S^\circ [C(s)] = 6.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr & S^\circ [C{O_2}(g)] = 210.0\,J\,{K^{ - 1}}mo{l^{ - 1}} \cr} $$
Assume that, the enthalpies and the entropies are temperature independent.
Answer
935
935k
- OR
935K
- OR
Explanation
Tin is obtained from cassiterite by reduction with coke and the balanced chemical reaction is
$$Sn{O_2}(s) + C(s)\buildrel {} \over \longrightarrow Sn(s) + C{O_2}(g)$$
Standard enthalpy of reaction,
$$\Delta H_{{R^\pi }}^o = (\Delta H_f^oC{O_2}(g)) - (\Delta H_f^oSn{O_2}(s))$$
$$\Delta H_{{R^\pi }}^o = - 394 - ( - 581) \Rightarrow 187$$
Standard entropy of reaction,
$$\Delta S_{{R^\pi }}^o = \Delta S_{\Pr oducts}^o - \Delta S_{{\mathop{\rm Re}\nolimits} ac\tan ts}^o$$
$$\Delta S_{{R^\pi }}^o = [S_{}^o(Sn(s)) + S_{}^o(C{O_2}(g))] - S_{}^o(Sn{O_2}(s)) + S_{}^o(C(s))]$$
$$\Delta S_{{R^\pi }}^o = [52 + 210] - [56 + 6]$$
$$ \Rightarrow $$ 200 JK$$-$$1 mol$$-$$1
We know that, $$\Delta H^\circ = T\Delta S^\circ $$
$$ \therefore $$ $$T = {{\Delta H^\circ } \over {\Delta S^\circ }}$$
$$ = {{187 \times 1000} \over {200}} \Rightarrow 935K$$
For the reaction to be spontaneous, the temperature should be greater than 935 K.
$$Sn{O_2}(s) + C(s)\buildrel {} \over \longrightarrow Sn(s) + C{O_2}(g)$$
Standard enthalpy of reaction,
$$\Delta H_{{R^\pi }}^o = (\Delta H_f^oC{O_2}(g)) - (\Delta H_f^oSn{O_2}(s))$$
$$\Delta H_{{R^\pi }}^o = - 394 - ( - 581) \Rightarrow 187$$
Standard entropy of reaction,
$$\Delta S_{{R^\pi }}^o = \Delta S_{\Pr oducts}^o - \Delta S_{{\mathop{\rm Re}\nolimits} ac\tan ts}^o$$
$$\Delta S_{{R^\pi }}^o = [S_{}^o(Sn(s)) + S_{}^o(C{O_2}(g))] - S_{}^o(Sn{O_2}(s)) + S_{}^o(C(s))]$$
$$\Delta S_{{R^\pi }}^o = [52 + 210] - [56 + 6]$$
$$ \Rightarrow $$ 200 JK$$-$$1 mol$$-$$1
We know that, $$\Delta H^\circ = T\Delta S^\circ $$
$$ \therefore $$ $$T = {{\Delta H^\circ } \over {\Delta S^\circ }}$$
$$ = {{187 \times 1000} \over {200}} \Rightarrow 935K$$
For the reaction to be spontaneous, the temperature should be greater than 935 K.
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