PCl5g- PCl3g -Cl2g at 298 K- Kc-1-8- 10-7- What is - G0 for the reaction-

The correct option is D 38.484 kJ/mol Δ G^0= RTlnK #160; #160; #160; #160; #160; Δ G^0= 8.314×298× ln(1.8×10^ 7) #160; #160; # ...

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Byju's Answer

Standard XII

Chemistry

Enthalpy of Combustion

PCl5g→ PCl3g ...

Question

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$ at 298 K, $K_{c} = 1.8 \times 10^{- 7} .$ What is $Δ G^{0}$ for the reaction?

30.484 kJ/mol

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28.484 kJ/mol

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24.484 kJ/mol

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38.484 kJ/mol

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Solution

The correct option is D 38.484 kJ/mol
$Δ G^{0} = - R T l n K$

$Δ G^{0} = - 8.314 \times 298 \times l n (1.8 \times 10^{- 7})$

$Δ G^{0} = 38.48 k J / m o l$

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Similar questions

Q. Calculate:

Δ G^{0}

b) the equilibrium constant for the formation of

N O_{2}

from NO and

O_{2}

at 298 K.

N O (g) + \frac{1}{2} O_{2} (g) ⇌ N O_{2} (g)

where

Δ_{f} G^{0} (N O_{2}) = 52.0

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Δ_{f} G^{0} (N O) = 87.0

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Δ_{f} G^{0} (O_{2}) = 0

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Q. For the equilibirium,

P {C l}_{5} (g) ⇌ P {C l}_{3} (g) + {C l}_{2} (g)

K_{c} = 1.8 \times 10^{- 7}

Calculate

Δ G^{o}

for the reaction

(R = 8.314 J K^{- 1} {m o l}^{- 1})

The equilibrium constant for the reaction given below is $2.0 \times 10^{- 7} a t 300 K$ .
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$ . The value of $| S^{\circ} |$ is $- x J m o l^{- 1} K^{- 1}$ . Calculate the value of $x$ if $Δ H^{\circ} = 28.4 k J m o l^{- 1}$ (Give your answer upto one deimal only)

Q. For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given. What is the nature of reaction?

I_{2} + H_{2} S \to 2 H I + S

△ G_{f}^{o} (H I) = 1.8 k J m o l^{- 1}, △ G_{f}^{o} (H_{2} S) = 33.8 k J m o l^{- 1}

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
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PCl5(g)→PCl3(g)+Cl2(g) at 298 K, Kc=1.8×10−7. What is ΔG0 for the reaction?

The correct option is D 38.484 kJ/molΔG0=−RTlnK ΔG0=−8.314×298×ln(1.8×10−7) ΔG0=38.48 kJ/mol

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$ at 298 K, $K_{c} = 1.8 \times 10^{- 7} .$ What is $Δ G^{0}$ for the reaction?

The correct option is D 38.484 kJ/mol
$Δ G^{0} = - R T l n K$

$Δ G^{0} = - 8.314 \times 298 \times l n (1.8 \times 10^{- 7})$

$Δ G^{0} = 38.48 k J / m o l$