At a certain temperature, a 21.0-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 13.5 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients)?

SO2 +N2O<--->SO3+N2
Kc =  concentration   of  product /concentration  of  reactant
(SO3)(N2/(SO2)(N2O)
Find  the  concentration  of  each reactant  and  products
concentration =number  of  moles/  volume  in  liters
 moles  of SO3=  3.5/80=0.044moles
SO2=4.6/64=0.072 moles
N2=13.5/28=0.483  moles
N2O=0.98/44=0.022 moles
concentration  is  therefore=
SO3=0.044/21=  2.095  x  10^-3
SO2=0.072 /21=3.43 x  10^-3
N2=0.483/21=0.023
N2O=0.022/22= 1.048 x 10^-3

Kc   is  therefore  ={ (2.095 x10^-3)(0.023)}/ {(3.43 x10^-3)(1.048  x10^-3)}=13.40

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

The answers is a nuclear changes