For the synthesis of ammonia at 300 K - N2 g -3H2 g - 2NH3 g Calculate the value of - G0 - N2 H2 NH3 - H0fKcal-mole 00-10S0 Cal-K-mole 403045

The correct option is A 8 Kcal Δ H^0_R=2×Δ H^0_f ( NH_3 ) 3×Δ H^0_f ( H_2 ) Δ H^0_f ( N_2 )= 20 Kcal Δ S^0 _R=2× S^0_ ( NH_3 ) 3× S^0_ ( H_2 ) S^0_ ...

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Byju's Answer

Standard XII

Physics

Alpha Decay

For the synth...

Question

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

-8 Kcal

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-12 Kcal

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-16 Kcal

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-20 Kcal

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Solution

The correct option is A -8 Kcal
$Δ H_{R}^{0} = 2 \times Δ H_{f (N H_{3})}^{0} - 3 \times Δ H_{f (H_{2})}^{0} - Δ H_{f (N_{2})}^{0} = - 20 K c a l$
$Δ S_{R}^{0} = 2 \times S_{(N H_{3})}^{0} - 3 \times S_{(H_{2})}^{0} - S_{(N_{2})}^{0}$
$= 2 \times 45 - 3 \times 30 - 40 = - 40 C a l / k$
$Δ G^{0} = Δ H^{0} - T Δ S^{0}$
$= - 20 - \frac{300 \times (- 40)}{1000} = - 20 + 12 = - 8 K c a l$

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Similar questions

N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g); Δ H = - 22 k c a l

Activation energy,

E_{a}

for the given reaction is

70 k c a l

. Find the activation energy for

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

k c a l

Q. For the reaction

A_{(l)} \to 2 B_{(g)}

Δ U = 2.1 kcal, Δ S = 20 cal/K

at 300 K
Hence

Δ G

k c a l

Q. For the reaction,

X_{2} O_{4 (l)} \to 2 X O_{2 (g)} Δ U = 2.1 k c a l, Δ S = 20 c a l K^{- 1} a t 300 K

Hence,

Δ G

Q. Determine the amount of heat (in kcal) given off at constant volume when

0.5 m o l

N_{2}

1.5

mol

H_{2}

reacted ording to equation at

300 K

N_{2} (g) + 3 H_{2} (g) \to {2 N H}_{3} (g); Δ_{r} H_{300} = - 380 k c a l / m o l

(Given: R=2 cal/ mol-K)

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) + 22

kcal. The activation energy for the forward reaction

50

kcal. What is the activation energy for the backward reaction?

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

For the synthesis of ammonia at 300 K :N2(g)+3H2(g)→2NH3(g)Calculate the value of ΔG0 ?N2H2NH3ΔH0_{f}(Kcal/mole)$00-10S0(Cal/K−mole)403045

The correct option is A -8 KcalΔH0R=2×ΔH0f(NH3)−3×ΔH0f(H2)−ΔH0f(N2)=−20KcalΔS0R=2×S0(NH3)−3×S0(H2)−S0(N2)=2×45−3×30−40=−40Cal/kΔG0=ΔH0−TΔS0=−20−300×(−40)1000=−20+12=−8Kcal

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