Now suppose a reaction vessel is filled with 0.505 atm of ammonia (NH3) and 3.46

Question

Now suppose a reaction vessel is filled with 0.505 atm of ammonia (NH3) and 3.46 atm of nitrogen (N2) at 839. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of N tend to rise or fall? fall Is it possible to reverse this tendency by adding H,? In other words, if you said the pressure of N2 will tend to rise, can that be changed to a tendency to fall by adding H2? Similarly, if you said the pressure of N2 will tend to fall, can that be changed to a tendency to rise by adding H2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H needed to reverse it. Round your answer to 2 significant digits. yes no atm

Explanation / Answer

a)

if no pressure of H2 is present, then H2 must be produced

therefore, NH3 forms both, Ne and H2

N2 must rise

b)

yes, we could, via addition of H2

then

reaction will be:

2NH3 = N2 + 3H2

dG 34

K = exp(-dG/(RT))

K = exp(-34000/(8.314*(839+273))

K = 0.02528

now..

Q = [N2][H2]^3 / ([NH3]^2)

find q required

0.02528 = (3.40)(H2)^3 / (0.505^2)

H2^3 = 0.02528/3.40*(0.505^2)

H2 = (0.001896)^(1/3)  

P-H2 = 0.1237 atm

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