What is the change in enthalpy when 9.00 mol of sulfur trioxide decomposes to sulfur dioxide and oxygen gas

-891kj

Explanation:

We have the chemical equation as:

2SO2(g) + O2(g) 2SO3; ∆H° = 198kj

If sulphur trioxide decomposes we have opposite reaction. Enthalpy of decomposition will not be positive for forward reaction.

2 moles of SO3 ∆H° = -198Kj

1 mole = -198/2

9 moles = -198/2 x 9

= -891kj

Therefore change in enthalpy when 9 moles is decomposes is -891kj.

the equation that correctly represent the reaction for formation of ammonia is

n2+ 3h2 → 2nh3 (answer d)

1 mole of nitrogen gas (n2) react with 3 moles of hydrogen gas (h2) to form 2 moles of ammonia ( nh3). this is in a process known as haber process were iron is used as a catalyst and reaction take place that a higher temperature and pressure. the process is exothermic hence energy is released.

it's all four options.