Amixture consisting of 0.250 m n2(g) and 0.500 m h2(g) reaches equilibrium accordin pavlovic (rp32269) – homework 2 – campion – (48940) 11 to the equation n2(g) + 3 h2(g) → 2 nh3(g). at equilibrium, the concentration of ammonia is 0.150 m. calculate the concentration of n2(g) at equilibrium.

0.175M

Explanation:

To solve this problem, an ICE chart is used. The acronym ICE stands for initial, change and equilibrium.

At the initial stage of the reaction, the concentration of the product is at zero while that of the reactants are used as given. Then the change is constant across all boards and this can be represented with any variable say x for instance.

The change in concentration of the reactants is termed negative. The number of moles of the reactants and products are taken into consideration however. Hence we multiply the change in concentration by the number of moles. Let us solve the problem as shown below:

N2. H2. NH3

I 0.25. 0.5. 0

C. -x. -3x. 2x

E. 0.25-x. 0.5-3x 0.15

From the table, we can see that 2x equals 0.15

2x = 0.15

x = 0.15/2 = 0.075

Now in the column of Hydrogen, the equilibrium concentration of nitrogen is 0.25 - x and that is 0.25 - 0.075 = 0.175M

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation:

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 :