Question

Find out whether it is possible to reduce MgO using carbon at 298 K. If not, at what temperature, it becomes spontaneous for reaction, MgO(s) + C(s) → Mg(s) + CO(g), Given : ΔH° = + 491.18 kJ $mol^{-1}$ and ΔS° = + 197.67 J$K^{-1}$ $mol^{-1}$ at 298 K.

Answer 1

The given chemical reaction between MgO and C is as follows,

$MgO(s)\quad +\quad C(s)\quad \rightarrow \quad Mg(s)\quad +\quad CO(g)\\ { \Delta H }^{ o }\quad =\quad +\quad 491.18\quad kJ\quad mo{ l }^{ -1 }$

$MgO(s)\quad +\quad C(s)\quad \rightarrow \quad Mg(s)\quad +\quad CO(g)$

${ \Delta H }^{ o }\quad =\quad +\quad 491.18\quad kJ\quad mo{ l }^{ -1 }$

${ \Delta S }^{ o }\quad \quad =\quad 197.67\quad J{ K }^{ -1 }mo{ l }^{ -1 }$

${ \Delta G }^{ o }\quad =\quad { \Delta H }^{ o }\quad -\quad T\Delta S$

$=\quad 491.18\quad -\quad 298\quad \times \quad [\quad 197.67\quad \times 1{ 0 }^{ -3 }]$

$=\quad 432.27\quad kJ$

Thus reaction is non-spontaneous at 298.

For spontaneous nature

$\Delta { G }^{ 0 }\quad =\quad -ve$

$T\Delta { S }^{ 0 }\quad >\quad \Delta { H }^{ o }$

(Or)

$T\quad \times \quad [\quad 197.67\quad \times \quad { 10 }^{ -3 }]\quad >\quad 491.18$

$T\quad >\quad \frac { 491.18 }{ 197.67\quad \times \quad { 10 }^{ -3 } } \quad >\quad 2484.8\quad K$

Answer 2

Answer:

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