Calculate - f G - for NH 4 Cl - s at 300 K-Given- - f H - NH 4 Cl - s -314-5 kJ - mol - - r C p -0S - N - 2 g -lcirc-192 JK -1 mol -1 - - S - H - 2 g - hcirc -130-5 JK -1 mol -1S C l- 2g- kirc -233 JK -1 mol -1 - S - NH - 4 Cl s 99-5 JK -1 mol -1All given data at 300 K-A- -193-56 kJ mol -1B- -426-7 kJ mol -1C- -202-3 kJ mol -1D- None of these

The correct option is C 202.3 kJ mol^ 112N_2(g) + 2H_2(g) + 12Cl_2(g)→ NH_4Cl(s) Δ_fS^∘ (NH_4Cl, s) at 300 K = S_NH_4 Cl(s)^∘ [12S_N_2^∘ + 2S_H_2^∘ + ...

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Byju's Answer

Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

Calculate Δ f...

Question

Calculate $Δ_{f} G^{\circ}$ for $(N H_{4} C l, (s))$ at $300 K$ .
Given: $Δ_{f} H^{\circ} (N H_{4} C l, s) = - 314.5 k J / m o l : Δ_{r} C_{p} = 0$
$S_{N_{2} (g)}^{\circ} = 192 J K^{- 1} m o l^{- 1}; S_{H_{2} (g)}^{\circ} = 130.5 J K^{- 1} m o l^{- 1} S_{C l_{2} (g)}^{\circ} = 233 J K^{- 1} m o l^{- 1}; S_{N H_{4} C l (s)} 99.5 J K^{- 1} m o l^{- 1}$
All given data at $300 K$ .

- 193.56 k J m o l^{- 1}

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- 426.7 k J m o l^{- 1}

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- 202.3 k J m o l^{- 1}

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None of these

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Solution

The correct option is C $- 202.3 k J m o l^{- 1}$
$\frac{1}{2} N_{2} (g) + 2 H_{2} (g) + \frac{1}{2} C l_{2} (g) \to N H_{4} C l (s)$

$Δ_{f} S^{\circ} (N H_{4} C l, s) a t 300 K = S_{N H_{4} C l (s)}^{\circ} - [\frac{1}{2} S_{N_{2}}^{\circ} + 2 S_{H_{2}}^{\circ} + \frac{1}{2} S_{C l_{2}}^{\circ}] = - 374 J K^{- 1} m o l^{- 1} ∵ Δ_{r} C_{p} = 0 ∴ Δ_{f} f_{300}^{\circ} = Δ_{f} S_{300}^{\circ} = - 374 J K^{- 1} m o l^{- 1} Δ_{f} H_{300}^{\circ} = - 314.5 k J m o l^{- 1} Δ_{f} G_{300}^{\circ} = Δ_{f} H^{\circ} - 300 Δ S^{\circ} = - 314.5 - \frac{300 (- 374)}{1000} = - 202.3 k J m o l^{- 1}$

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Similar questions

Q. Calculate

Δ_{f} G^{\circ}

for

(N H_{4} C l, (s))

300 K

.
Given:

Δ_{f} H^{\circ} (N H_{4} C l, s) = - 314.5 k J / m o l : Δ_{r} C_{p} = 0

S_{N_{2} (g)}^{\circ} = 192 J K^{- 1} m o l^{- 1}; S_{H_{2} (g)}^{\circ} = 130.5 J K^{- 1} m o l^{- 1} S_{C l_{2} (g)}^{\circ} = 233 J K^{- 1} m o l^{- 1}; S_{N H_{4} C l (s)}^{\circ} = 99.5 J K^{- 1} m o l^{- 1}

All given data at

300 K

Q. Calculate

△_{f} G^{o}

k J m o l^{- 1}

for

(N H_{4} C l, s)

at 310 K.
Given :

△_{f} H^{o} (N H_{4} C l, s) = - 314.5 k J m o l^{- 1}; △_{r} C_{p} = 0

S_{N_{2} (g)}^{o} = 192 J K^{- 1} m o l^{- 1}; S_{H_{2} (g)}^{o} = 130.5 J K^{- 1} m o l^{- 1};

S_{C l_{2}}^{o} (g) = 233 J K^{- 1} m o l^{- 1}; S_{N H_{4} C l (s)}^{o} = 99.5 J K^{- 1} m o l^{- 1}

All given data are at 300 K.

Q. The

Δ_{f} G^{\circ}

for

(N H_{4} C l, s)

310 K

- x {kJ mol}^{- 1}

.
Given:

Δ_{f} H^{\circ} (N H_{4} C l, s) = - 314.5 {kJ mol}^{- 1}; Δ_{r} C_{P} = 0 S_{N_{2} (g)}^{\circ} = 192 {J K}^{- 1} {mol}^{- 1}; S_{H_{2} (g)}^{\circ} = 130.5 {J K}^{- 1} {mol}^{- 1} S_{C l_{2} (g)}^{\circ} = 223 {J K}^{- 1} {mol}^{- 1}; S_{N H_{4} C l (s)}^{\circ} = 99.5 {J K}^{- 1} {mol}^{- 1}

All given data at

300 K

. Find the magnitude of

x

Q. A reaction is at equilibrium at

153^{o} C

and the enthalpy change for the reaction is

42.6 k J m o l^{- 1} .

What will be the value of

△ S

Calculate the change in entropy for the following reaction
$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ .
Given $S_{C O_{2}}^{0} = 213.6 J K^{- 1} m o l^{- 1}$ , $S_{C O}^{0} = 297.6 J K^{- 1} m o l^{- 1}$ respectively.

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Calculate ΔfG∘ for (NH4Cl,(s)) at 300 K. Given: ΔfH∘(NH4Cl,s)=−314.5kJ/mol:ΔrCp=0 S∘N2(g)=192JK−1mol−1; S∘H2(g)=130.5JK−1mol−1S∘Cl2(g)=233JK−1mol−1;SNH4Cl(s)99.5JK−1mol−1 All given data at 300 K.