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The values of \Delta H \ and \ \Delta S for the reaction, C_{\left ( graphite \right )}+CO_{2}_{\left ( g \right )}\rightarrow 2CO_{\left ( g \right )}  are 170kJ and 170JK^{-1}, respectively. The reaction will be spontaneous at

  • Option 1)

    910K

  • Option 2)

    1110K

  • Option 3)

    510K

  • Option 4)

    710K

 

Answers (1)

 

Gibb's free energy (? G) -

\Delta G= \Delta H-T \Delta S
 

- wherein

\Delta G= Gibb's free energy

\Delta H= enthalpy of reaction

\Delta S= entropy

T= temperature

 

 For the reaction to be spontaneous \Delta G = -Ve

given \Delta G = 170\times 10^{3}J, \Delta S =170 JK^{-1}\:mol^{-1}

\Delta G is -Ve only when T\Delta S> \Delta H, which is possible only at T=1110K ie. reaction will be spontaneous

\therefore \Delta G=170\times 10^{3}-1110\times 170= -18700

P=\frac{nRT}{V}\\ n=\frac{m}{M}=\frac{6}{16}=\frac{3}{8}\:mole \\ T=129^oc=402K

V=.03m3

P\frac{nRT}{v}=\frac{3}{8}\times \frac{8.314\times 402}{.03}pa\\ =41648\:pa


Option 1)

910K

Incorrect Option

Option 2)

1110K

Correct Option

Option 3)

510K

Incorrect Option

Option 4)

710K

Incorrect Option

Posted by

subam

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