Chemistry, asked by sn0350251, 3 days ago

Subject Test < Jote You are attempting question 3 o For the reaction N2(g) + 3H2(g) + 2NH, (8), equilibrium pressures are P. = 2.6 am, Pa = 5.6 atm, and Px; – 1.2 atm . Calculate the Gibb’s free energy of the reaction at a temperature of 298K. (A) 1.43 kJ (B) 1.43x10- J (C) 1.41x103 J (D) 4.25x10' J niswer.​

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Answered by yatharthgupta285
0

Answer:

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For the reaction; N

2(g)

+3H

2(g)

⇌2NH

3(g)

.

At 400 K,K

p

=41 atm

−2

. Find the value of K

p

for each of the following reactions at the same temperature:

(i) 2NH

3(g)

⇌N

2(g)

+3H

2(g)

;

(ii)

2

1

N

2(g)

+

2

3

H

2(g)

⇌NH

3(g)

;

(iii) 2N

2(g)

+6H

2(g)

⇌4NH

3(g)

.

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Answer

(i) K

p

=

[NH

3

]

2

[H

2

]

3

[N

2

]

=[

[N

2

][H

2

]

3

[NH

3

]

2

]

−1

=(41)

−1

=

41

1

=0.024 atm

2

(∵

[N

2

][H

2

]

3

[NH

3

]

2

=41)

(ii) K

p

=

[H

2

]

3/2

[N

2

]

1/2

[NH

3

]

=[

[H

2

]

3

[N

2

]

[NH

3

]

2

]

1/2

=(41)

1/2

=6.4 atm

−1

(iii) K

p

=

[N

2

]

2

[H

2

]

6

[NH

3

]

4

=[

[N

2

][H

2

]

3

[NH

3

]

2

]

2

=(41)

2

=1.681×10

3

atm

−4

.

Explanation:

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Answered by krsankar659
0

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