Question

answer the attachment and then answer a one more question,

following reaction is performed at 298K.

$\sf 2NO(g)+O_2(g)\rightleftharpoons 2NO_2 (g)$

the standard free energy of formation of NO(g) is 86.6 kJ/mol at 298K. what is standard free energy of formation of NO_2(g) at 298K?​

Answer 1

answer for first Question:-

perimeter $\sf = 2(2\boldsymbol\alpha +2 \ cos \ 2 \boldsymbol\alpha)$

$\sf = 4(\boldsymbol\alpha + cos \ 2\boldsymbol\alpha)$

$\sf \dfrac{dP}{d\boldsymbol\alpha} = 4(1-2 \ sin \ 2\boldsymbol\alpha) = 0$

$\sf sin \ 2\boldsymbol\alpha = \dfrac{1}{2}$

$\sf 2\boldsymbol\alpha = \dfrac{\pi}{6} , \dfrac{5\pi}{6}$

$\sf \dfrac{d^2P}{d\boldsymbol\alpha^2} = -4 \ cos \ 2\boldsymbol\alpha$

for $\sf max(\boldsymbol\alpha) = \dfrac{\pi}{12}$

area $\sf = (2\boldsymbol\alpha) \times (2\ cos \ 2 \boldsymbol\alpha)$

$\sf = \dfrac{\pi}{6} \times 2 \times \dfrac{\sqrt{3}}{2}$

rationalise the denominator

$\sf = \dfrac{\pi}{6} \times \sqrt{3} \times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\sf = \dfrac{\pi \sqrt{3}^2}{6\sqrt{3}}$

$\boxed{\sf = \dfrac{{}^1 \cancel{3} \pi}{{}^2 \cancel{6} \sqrt{3}} = \dfrac{\pi}{2\sqrt{3}}}$

answer for second question:

$\sf ( \rm \Delta \sf G^{\circ} )_{reaction} = \bigg[ ( \rm \Delta \sf G^{\circ} )_{formation}\bigg]_{product}$

$\sf \implies -RT \ ln \ K_p = 2( \rm \Delta \sf G^{\circ} )_{NO_{2}} - 2( \rm \Delta \sf G^{\circ} )_{NO}$

$\sf \implies ( \rm \Delta \sf G^{\circ} )_{NO_{2}} = ( \rm \Delta \sf G^{\circ} )_{NO} - RT \ ln \ K_p$

$\sf \implies ( \rm \Delta \sf G^{\circ} ) = \dfrac{2\times 86600 - R(298) \ ln \ K_p}{2}$

$\boxed{\sf = 0.5(2\times 86600 -R(298) \ ln \ 1.6 \times 10^{12} )}$

Answer 2

Answer:

Hiii Damsildevil

thanks alot yrr

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