Question

If the standard heat of formation of ammonia gas at 25°C was found to be - 11.04 k cal. Calculate the heat of solution at 348 K. Given that the mean heat capacities of Nx H and NH; are 6.80.6.77 and 8.86 cal/degree/mol respectively. [Hint: No + H — NH; + 2 AC = (C) NH – } (9) N:- (9.) H. CN 3 2 i T = 25 + 273 = 298 K, T, = 348 K ​

Answer 1

Answer:

2

1

N

2

(g)+

2

3

H

2

(g)⟶NH

3

(g)

ΔH

298K

=∑(H

f

)

P

−∑(H

f

)

R

=(−11.02−0)=−11.02kcalmol

−1

T

2

−T

1

ΔH

2

−ΔH

1

=ΔC

P

773−298

ΔH

2

−(−11.02)

=(8.38−

2

1

×6.96−

2

3

×6.89)×10

−3

H2

=−13.6kcalmol

−1

Explanation:

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