Question

Then the value of ΔH of reaction, ΔG_1 and ΔG_2 (in kg) at temperature T_1 and T_2 respectively are?

Answer 1

ANSWER.

The value of ∆H of the reaction ∆G1 and ∆G2

( in kj) at temperature T1 and T2

=> 46.14 , -6.2 , - 14.3

option [ 1 ] is correct

EXPLANATION.

$\sf \to \: t_{1} \: = 50 {}^{0}c \: \: \: and \: \: \: k_{1} = 10 \\ \\ \sf \to \: t_{2} = 100 {}^{0}c \: \: \: \: and \: \: \: k_{2} \: = 100$

$\sf \to \: formula \: of \: \Delta \: h \: of \: reaction \\ \\ \sf \to \: log( \frac{ k_{2}}{ k_{1} } ) = \frac{ \Delta \: H }{2.303R} ( \frac{1}{ t_{1} } - \frac{1}{ t_{2}})$

$\sf \to \: log( \dfrac{100}{10} ) = \dfrac{ \Delta \: H }{2.303 \times 8.314} ( \dfrac{1}{323} - \dfrac{1}{373} ) \\ \\ \\ \sf \to \: log(10) = \frac{ \Delta \:H }{2.303 \times 8.314} ( \frac{50}{323 \times 373} ) \\ \\ \\ \sf \to \: \Delta \: H \: = \frac{2.303 \times 8.314 \times 323 \times 373}{50} \\ \\ \\ \sf \to \: \Delta \: H = 46136.57j \: = 46.14kj$

$\sf \to \: \Delta \: G \: = - 2.303 RT log(k) \\ \\ \sf \to \: at \: t_{1} = 50 {}^{0} = 323k \: \: and \: \: k_{1} = 10 \\ \\ \sf \to \: \Delta \: G = - 2.303 \times 8.314 \times 323 \times log(10) \\ \\ \sf \to \: \Delta \: G = - 6184.5j = 6.2kj$

$\sf \to \: \Delta \: G = - 2.303RT log(k) \\ \\ \sf \to \: at \: t_{2} = 100 {}^{0} c \: = 373k \: \: \: and \: \: \: \: k_{2} = 100 \\ \\ \sf \to \: \Delta \: G = - 2.303 \times 8.314 \times 373 \times log(100) \\ \\ \sf \to \: \Delta \: G = - 14283.7j = - 14.3kj$

Answer 2

$\sf \implies \:The \:value \:of \:H \:of \:the \:reaction \:G1 \:and \:G2\:$

$\sf \implies \:( in kj) \:at \:temperature \:T1 \:and \:T2\:$

$\sf \implies \: 46.14 \:, -6.2 \:, - \:14.3\:$

$\sf \implies \: option \:[ 1 ] \:is \:correct\:$