Solution:- Balanced equation is.... SO 3 (g) + N 2 (g) <-----> SO 2 (g) + N 2 O(

Question

Solution:- Balanced equation is....

SO3(g) + N2(g) <-----> SO2(g) + N2O(g)

moles of SO3 = 3.5 g x (1mol/80 g) = 0.044 mol SO3

moles of N2 = 24.1 g x (1mol/28g) = 0.86 mol N2

moles of SO2 = 4.6 g x (1mol/64 g) = 0.072 mol SO2

moles of N2O = 0.98 g x (1mol/44g) = 0.022 mol N2O

Divide the moles of each by the given volume, 20.0 L to get the initial concentrations.

[SO2] = 0.072 mol/20.0L = 0.0036 M

[N2O] = 0.022 mol/20.0L = 0.0011 M

[SO3] = 0.044 mol/20.0L = 0.0022 M

[N2] = 0.86 mol/20.0L = 0.043 M

Equilibrum constant, Kc = [SO2][N2O]/[SO3][N2]

Kc = [(0.0036)(0.0011)]/[(0.0022)(0.043)]

Kc = 0.042

Explanation / Answer

t a certain temperature, a 20.0L contains holds four gases in equilibrium. Their masses are: .5 g SOs. 4.6 g SO2, 24.1 g N2. and 0.98 g N2O. What is the value of the equilibrium constant at this temperature for the reaction of SO2 with N20 to form SOs and N2 (balanced with lowest whole-number coefficients)?

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