Question

In: Chemistry

You wish to monitor the progress of a reaction using pressures. Suppose 17.88g N2 (28.02g/mol) and...

You wish to monitor the progress of a reaction using pressures. Suppose 17.88g N2 (28.02g/mol) and 1.328g H2 (2.016g/mol) are initially present in a 25.0L reaction vessel at 124 degrees celsius. Nitrogen and hydrogen can react to form ammonia (17.04g/mol) according to the reaction shown. At a given point, the total pressure (the sume of N2, H2 and Nh3 present) is 1.62atm. What percentage of the initial N2 has reacted?

Expert Solution

N2 (g) + 3 H2 (g) -------> 2 NH3 (g)

initially for N2 = 17.88/28.02 = 0.6381 moles

initial pressure of N2, PV = nRT

P * 25 = 0.6381 * 0.0821 * (124+273)

P = 0.8391 atm

initially for H2 = 1.328/2.016 = 0.6587 moles

initial pressure of H2, PV = nRT

P * 25 = 0.6587 * 0.0821 * (124+273)

P = 0.8588 atm

N2 (g) + 3 H2 (g) -------> 2 NH3 (g)

initially 0.8391 0.8588 0

at equi (0.8391 - x) (0.8588-3x) 2x

But it is given,

(0.8391 - x)+(0.8588-3x)+2x = 1.62

x = 0.03895

Percentage of N2 reacted = 0.03895/0.08391*100 = 46.42%

queen_honey_blossom answered 2 years ago

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