What is - rG kJ-mole for synthesis of ammonia at 298 K at following sets of partial pressure - N2g-3H2g- 2NH3g- - rG-33 kJ-mol-Take R-8-3 J-K mole- log 2-0-3- log 3-0-48-Gas - N2 H2 NH3Pressure atm - 1 3 0-02

The correct option is C 60.5 Δ_rG = Δ_rG^0 + RTlnQ_r #160; #160; , where Q_r is the reaction Quotient.Now, Q_r = p_NH_3^2p_H_2^3 × p_ ...

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Byju's Answer

Standard XII

Chemistry

Rate of Reaction

What is Δ r...

Question

What is $Δ_{r} G (k J / m o l e)$ for synthesis of ammonia at 298 K at following sets of partial pressure :
$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); Δ_{r} G^{\circ} = - 33 k J / m o l$

[Take $R = 8.3 J / K m o l e, log 2 = 0.3; log 3 = 0.48$ ]
Gas : $N_{2}$ $H_{2}$ $N H_{3}$
Pressure (atm) : 1 3 0.02

+ 6.5

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- 6.5

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+ 60.5

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- 60.5

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Solution

The correct option is C $- 60.5$
$Δ_{r} G = Δ_{r} G^{0} + R T l n Q_{r}$ , where $Q_{r}$ is the reaction Quotient.
Now, $Q_{r} = \frac{p_{N H_{3}}^{2}}{p_{H_{2}}^{3} \times p_{N_{2}}} = \frac{{0.02}^{2}}{3^{3} \times 1}$
Substituting values of $Δ_{r} G^{0}, Q_{r}, R$ and $T$ , we get, $Δ_{r} G = - 60.5 k J / m o l$ .

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Similar questions

Q. 1 mole

N_{2}

and 3 mol

H_{2}

are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained.

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

. The equilibrium constant

K_{p}

for dissociation of

N H_{3}

is:

1

mole of

N_{2}

and

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moles of

H_{2}

are placed in a closed container at a pressure of

4

atm. The pressure falls to

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atm at the same temperature when the following equilibrium is attained:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

The equilibrium constant

K_{p}

for dissociation of

N H_{3}

is:

Q. In the following, equilibrium partial pressure of

N H_{3}, N_{2}

and

H_{2}

gasses are 4, 1 and 2 atm respetively at 300 K. what is value of

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N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

Q. In a mixture of

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Calculate the equilibrium constant of the reaction using the equation.

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) .

Q. For the synthesis of ammonia at 300 K :

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)

Calculate the value of

Δ G^{0}

	$N_{2}$	$H_{2}$	$N H_{3}$
$Δ H^{0}$ _{f}(Kcal/mole)$	0	0	-10
$S^{0} (^{C} a l / K - m o l e)$	40	30	45

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What is ΔrG(kJ/mole) for synthesis of ammonia at 298 K at following sets of partial pressure :N2(g)+3H2(g)⇌2NH3(g);ΔrG∘=−33kJ/mol[Take R=8.3J/Kmole,log2=0.3;log3=0.48]Gas : N2 H2 NH3Pressure (atm) : 1 3 0.02

The correct option is C −60.5ΔrG=ΔrG0+RTlnQr, where Qr is the reaction Quotient.Now, Qr=p2NH3p3H2×pN2=0.02233×1Substituting values of ΔrG0,Qr,R and T, we get, ΔrG=−60.5kJ/mol.