What effect will spraying liquid water into the equilibrium given below have if nh3 is far more soluble in water than is n2 or h2?
n2(g) + 3h2(g) 2nh3(g)
a) more nh3(g) will form.
b) more h2(g) will form.
c) less nh3(g) will form.
d) this will not affect the system.
e) more n2(g) will form.

A) More NH3(g) will form.

Explanation:

According to the Le Chatellier Principle if we remove product from a reaction which is in equilibrium, then the effect will to produce more product in order to replace what has been removed.

In this reaction also , on spraying liquid water, Ammonia being water soluble will be consumed. Hence, The reaction will shift in forward direction leading to more formation of Ammonia (NH3).

Hence option A will be correct .

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :