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CHXI07:EQUILIBRIUM

314875 If for the heterogeneous equilibrium
$CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)$ , $K = 1$ $a t$ $1atm$ pressure, the temperature is given by

1

T = \frac{Δ S^{^{\circ}}}{Δ H^{^{\circ}}}

2

T = \frac{Δ H^{^{\circ}}}{Δ S^{^{\circ}}}

3

T = \frac{Δ G^{^{\circ}}}{R}

4

T = \frac{Δ G^{^{\circ}}}{Δ H^{^{\circ}}}

Explanation:

Condition for equilibrium obtained from the relation,
$Δ G^{\circ}$ = $- R T \ln K = - R T \ln 1 = 0$
$Δ H^{\circ} - T Δ S^{\circ} = 0$
$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$

CHXI07:EQUILIBRIUM

314876 Match Column I with Column II and choose the correct combination from the options given
supporting img

1 A - S, B - Q,R, C - Q, D - P

2 A - Q, B - S, C - R, D - P

3 A - P, B - R, C - Q , D - S

4 A - S, B - P, C - R, D - Q

Explanation:

(A-S); At equilibrium, Q = K.
(B-Q,R); When $Q < K$ , reaction proceeds in forward direction.
(C-Q); When $Q > K$ , reaction proceeds in backward direction.
(D-P); When $K >> 1$ , reaction is about to complete.

CHXI07:EQUILIBRIUM

314877 If equilibrium constant is $26 \times 10^{8}$ at $298 K$ then find the value of $Δ G^{^{\circ}}$

1

- 48.014 kJ

2

- 63.2 kJ

3

- 23.3 kJ

4

+ 40 kJ

Explanation:

$Δ G^{\circ}$ is related to thermodynamic equilibrium constant as
$= 2.303 RT \log K$
$= - 2.303 \times 8.314 \times 298 \log (2.6 \times 10^{8})$
$= - 48014.56 J$
The answer should be $48.014 kJ$ .

CHXI07:EQUILIBRIUM

314878 Reactions that have standard free energy changes less than zero always have equilibrium constant equal to

1 unity

2 greater than unity

3 less than unity

4 zero

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K_{C}$
when $Δ G^{\circ} < 0$ , i.e. - ve, then $K_{C} > 1$ In this case, forward reaction is feasible showing thereby a large concentration of products.

CHXI07:EQUILIBRIUM

314875 If for the heterogeneous equilibrium
$CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)$ , $K = 1$ $a t$ $1atm$ pressure, the temperature is given by

1

T = \frac{Δ S^{^{\circ}}}{Δ H^{^{\circ}}}

2

T = \frac{Δ H^{^{\circ}}}{Δ S^{^{\circ}}}

3

T = \frac{Δ G^{^{\circ}}}{R}

4

T = \frac{Δ G^{^{\circ}}}{Δ H^{^{\circ}}}

Explanation:

Condition for equilibrium obtained from the relation,
$Δ G^{\circ}$ = $- R T \ln K = - R T \ln 1 = 0$
$Δ H^{\circ} - T Δ S^{\circ} = 0$
$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$

CHXI07:EQUILIBRIUM

314876 Match Column I with Column II and choose the correct combination from the options given
supporting img

1 A - S, B - Q,R, C - Q, D - P

2 A - Q, B - S, C - R, D - P

3 A - P, B - R, C - Q , D - S

4 A - S, B - P, C - R, D - Q

Explanation:

(A-S); At equilibrium, Q = K.
(B-Q,R); When $Q < K$ , reaction proceeds in forward direction.
(C-Q); When $Q > K$ , reaction proceeds in backward direction.
(D-P); When $K >> 1$ , reaction is about to complete.

CHXI07:EQUILIBRIUM

314877 If equilibrium constant is $26 \times 10^{8}$ at $298 K$ then find the value of $Δ G^{^{\circ}}$

1

- 48.014 kJ

2

- 63.2 kJ

3

- 23.3 kJ

4

+ 40 kJ

Explanation:

$Δ G^{\circ}$ is related to thermodynamic equilibrium constant as
$= 2.303 RT \log K$
$= - 2.303 \times 8.314 \times 298 \log (2.6 \times 10^{8})$
$= - 48014.56 J$
The answer should be $48.014 kJ$ .

CHXI07:EQUILIBRIUM

314878 Reactions that have standard free energy changes less than zero always have equilibrium constant equal to

1 unity

2 greater than unity

3 less than unity

4 zero

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K_{C}$
when $Δ G^{\circ} < 0$ , i.e. - ve, then $K_{C} > 1$ In this case, forward reaction is feasible showing thereby a large concentration of products.

CHXI07:EQUILIBRIUM

314875 If for the heterogeneous equilibrium
$CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)$ , $K = 1$ $a t$ $1atm$ pressure, the temperature is given by

1

T = \frac{Δ S^{^{\circ}}}{Δ H^{^{\circ}}}

2

T = \frac{Δ H^{^{\circ}}}{Δ S^{^{\circ}}}

3

T = \frac{Δ G^{^{\circ}}}{R}

4

T = \frac{Δ G^{^{\circ}}}{Δ H^{^{\circ}}}

Explanation:

Condition for equilibrium obtained from the relation,
$Δ G^{\circ}$ = $- R T \ln K = - R T \ln 1 = 0$
$Δ H^{\circ} - T Δ S^{\circ} = 0$
$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$

CHXI07:EQUILIBRIUM

314876 Match Column I with Column II and choose the correct combination from the options given
supporting img

1 A - S, B - Q,R, C - Q, D - P

2 A - Q, B - S, C - R, D - P

3 A - P, B - R, C - Q , D - S

4 A - S, B - P, C - R, D - Q

Explanation:

(A-S); At equilibrium, Q = K.
(B-Q,R); When $Q < K$ , reaction proceeds in forward direction.
(C-Q); When $Q > K$ , reaction proceeds in backward direction.
(D-P); When $K >> 1$ , reaction is about to complete.

CHXI07:EQUILIBRIUM

314877 If equilibrium constant is $26 \times 10^{8}$ at $298 K$ then find the value of $Δ G^{^{\circ}}$

1

- 48.014 kJ

2

- 63.2 kJ

3

- 23.3 kJ

4

+ 40 kJ

Explanation:

$Δ G^{\circ}$ is related to thermodynamic equilibrium constant as
$= 2.303 RT \log K$
$= - 2.303 \times 8.314 \times 298 \log (2.6 \times 10^{8})$
$= - 48014.56 J$
The answer should be $48.014 kJ$ .

CHXI07:EQUILIBRIUM

314878 Reactions that have standard free energy changes less than zero always have equilibrium constant equal to

1 unity

2 greater than unity

3 less than unity

4 zero

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K_{C}$
when $Δ G^{\circ} < 0$ , i.e. - ve, then $K_{C} > 1$ In this case, forward reaction is feasible showing thereby a large concentration of products.

CHXI07:EQUILIBRIUM

314875 If for the heterogeneous equilibrium
$CaC O_{3} (s) ⇌ CaO (s) + C O_{2} (g)$ , $K = 1$ $a t$ $1atm$ pressure, the temperature is given by

1

T = \frac{Δ S^{^{\circ}}}{Δ H^{^{\circ}}}

2

T = \frac{Δ H^{^{\circ}}}{Δ S^{^{\circ}}}

3

T = \frac{Δ G^{^{\circ}}}{R}

4

T = \frac{Δ G^{^{\circ}}}{Δ H^{^{\circ}}}

Explanation:

Condition for equilibrium obtained from the relation,
$Δ G^{\circ}$ = $- R T \ln K = - R T \ln 1 = 0$
$Δ H^{\circ} - T Δ S^{\circ} = 0$
$T = \frac{Δ H^{\circ}}{Δ S^{\circ}}$

CHXI07:EQUILIBRIUM

314876 Match Column I with Column II and choose the correct combination from the options given
supporting img

1 A - S, B - Q,R, C - Q, D - P

2 A - Q, B - S, C - R, D - P

3 A - P, B - R, C - Q , D - S

4 A - S, B - P, C - R, D - Q

Explanation:

(A-S); At equilibrium, Q = K.
(B-Q,R); When $Q < K$ , reaction proceeds in forward direction.
(C-Q); When $Q > K$ , reaction proceeds in backward direction.
(D-P); When $K >> 1$ , reaction is about to complete.

CHXI07:EQUILIBRIUM

314877 If equilibrium constant is $26 \times 10^{8}$ at $298 K$ then find the value of $Δ G^{^{\circ}}$

1

- 48.014 kJ

2

- 63.2 kJ

3

- 23.3 kJ

4

+ 40 kJ

Explanation:

$Δ G^{\circ}$ is related to thermodynamic equilibrium constant as
$= 2.303 RT \log K$
$= - 2.303 \times 8.314 \times 298 \log (2.6 \times 10^{8})$
$= - 48014.56 J$
The answer should be $48.014 kJ$ .

CHXI07:EQUILIBRIUM

314878 Reactions that have standard free energy changes less than zero always have equilibrium constant equal to

1 unity

2 greater than unity

3 less than unity

4 zero

Explanation:

$Δ G^{\circ} = - 2.303 RT \log K_{C}$
when $Δ G^{\circ} < 0$ , i.e. - ve, then $K_{C} > 1$ In this case, forward reaction is feasible showing thereby a large concentration of products.

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