Question

Suppose a 250 ml flask is filled with 1.0 mol of NO and 0.70 mol of NO . The following reaction becomes possible: NO3(8) + NO(g) - 2NO(8) The equilibrium constant K for this reaction is 5.78 at the temperature of the flask. Calculate the equilibrium molarity of NO3. Round your answer to two decimal places. Ом 3 ? Expiat tion Check 20:30 MCGHE AB

Answer 1

For reaction

NO3(g) + NO(g) 2 2 NO2(g)

given equilibrium constant K = 5.78

volume of container = 250 mL

initial moles of NO3 = 1.0 mol

initial moles of NO = 0.7 mol

Step 1.

calculation of initial molarity of reactants

[NO3] = no of moles of NO3/volume of container = 1 mol/250 mL = 1 mol/(250mL/1000mL)L = 1000molL-1/250 = 4 molL-1

[NO3] = 4 M

similiarly [NO] = 0.7*1000/250 = 2.8 M

Step 2.

using ICE method to calculate the equilibrium concentration of reactants and products

	NO3(g)	NO(g)	2 NO2
I (initially)	4	2.8	0
C (change)	-x	-x	2x	Reactants’ loss is gained as products
E (equilibrium)	4-x	2.8-x	2x

K = recatants/ products = [NO2]2/[NO3][NO] = (2x)2/(4-x)(2.8-x) = 5.78

4x2/(11.2-6.8x+x2) = 5.78;

1.78x2-39.30x+64.73 = 0

This is of following type

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using the formula

x = 1.79 and 20.28

only x = 1.79 is meaningful and acceptable solution.

therefore at equilibrium [NO3] = 4-x = 4-1.79 = 2.21 M