Question

For the following equilibrium: N_2(g) + H_2(g) rightarrow NH_3(g) delta H= -386 kJ/mole Predict the direction the equilibrium will shift if: N_2 is added? H_2 is removed? NH_3 is added? NH_3 is removed? the volume of the container is decreased? the pressure is increased by adding Argon gas? the reaction is cooled? equal number of moles of H_2 and NH_3 are added? a catalyst is added The equilibrium constant for the following reaction is 5.0 at 400 degree C. CO_(g) + H_2O_(g) reversible CO_2(g) + H_2(g) Determine the direction of the reaction if the following amount (in moles) of each compound is placed in a 1.0 L flask. The equilibrium concentrations of [O_2] = 0.21 Molar and [O_3] = 6.0 times 10^-8 Molar, calculate the value of K_c. O_3(g) reversible O_2(g)

Answer 1

6.

since dH is negative, this must be exothermic, meaning it gives off energy.

a)

if we add N2, a reactants, the shift goes toward product/right side

b)

if we remove H2, reactants, the shift goes toward reactants/left side

c)

if we add NH3, a product, the shift goes toward reactants/left ht side

d)

if we remove NH3, a product, the shift goes toward products/right side

e)

if volume is decreased, this favours the least mol formation, that is, NH3, products/right

f)

IF P increases, then this favours least mol formation that is, NH3, products/right

g)

if we cool down, the shift goes toward more products, since the shift goes toward heat production, i.e. products

h)

if we add H2 and NH3, we will actually have a higher effect on H2, since it has a coefficient of 3H2, whereas 2NH3... this shift goes to products

i) catalyst has no effect in equilibrium, only rate

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