The -fG- for NH4Cl- s at 310 K is -x kJ mol-1-Given- -fH- NH4Cl- s - - 314-5 kJ mol-1- -rCP - 0 SN2g- - 192 J K-1 mol-1 - SH2g- - 130-5 J K-1 mol-1SCl2g- - 223 J K-1mol-1 - SNH4Cls- - 99-5 J K-1 mol-1All given data at 300 K- Find the magnitude of x-

Δ_fS^∘(NH_4Cl, s) at 300 K = S_NH_4Cl(s)^∘ [12S_N_2^∘ + 2S_H_2^∘ + 12S_Cl_2^∘ ] = 369 J K^ 1 mol^ 1 ∵Δ_rC_p = 0 ∴Δ_fS_310^∘ = Δ_fS_300^∘ = 369 J K^ 1 ...

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Byju's Answer

Standard XII

Chemistry

Heat Capacity

The ΔfG∘ for ...

Question

The $Δ_{f} G^{\circ}$ for $(N H_{4} C l, s)$ at $310 K$ is $- x {kJ mol}^{- 1}$ .
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 $x$ .

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Solution

$Δ_{f} S^{\circ} (N H_{4} C l, s) a t 300 K = S_{N H_{4} C l (s)}^{\circ} - [\frac{1}{2} S_{N_{2}}^{\circ} + 2 S_{H_{2}}^{\circ} + \frac{1}{2} S_{C l_{2}}^{\circ}] = - 369 {J K}^{- 1} {mol}^{- 1} ∵ Δ_{r} C_{p} = 0 ∴ Δ_{f} S_{310}^{\circ} = Δ_{f} S_{300}^{\circ} = - 369 {J K}^{- 1} {mol}^{- 1} Δ_{f} H_{310}^{\circ} = Δ_{f} H_{300}^{\circ} = - 314.5 {kJ K}^{- 1} {mol}^{- 1} By using formula Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} Δ_{f} G_{310}^{\circ} = Δ_{f} H^{\circ} - 310 \times Δ S^{\circ} = - 314.5 - \frac{310 \times (- 369)}{1000} = - 200.11 {kJ mol}^{- 1}$
Hence, $x = 200.11$

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Similar questions

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. 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. 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)} 99.5 J K^{- 1} m o l^{- 1}

All given data at

300 K

Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place.

At $298 K$ : $Δ_{f} H^{\circ} (S n O_{2} (s)) = - 581.0 k J m o l^{- 1}, Δ_{f} H^{\circ} (C O_{2} (g)) = - 394.0 k J m o l^{- 1}$
$S^{\circ} (S n O_{2} (s)) = 56.0 J K^{- 1} m o l^{- 1}, S^{\circ} (S n (s)) = 52.0 J K^{- 1} m o l^{- 1}, S^{\circ} (C (s)) = 6.0 J K^{- 1} m o l^{- 1}, S^{\circ} (C O_{2} (g)) = 210.0 J K^{- 1} m o l^{- 1}$
Assume that the enthalpies and the entropies are temperature independent.

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The ΔfG∘ for (NH4Cl,s) at 310 K is −x kJ mol−1. Given: ΔfH∘(NH4Cl,s)=−314.5 kJ mol−1;ΔrCP=0S∘N2(g)=192 J K−1 mol−1; S∘H2(g)=130.5 J K−1 mol−1S∘Cl2(g)=223 J K−1mol−1; S∘NH4Cl(s)=99.5 J K−1 mol−1 All given data at 300 K. Find the magnitude of x.