Question

The following reaction is performed at a certain temperature and these data are acquired. Please use the reaction and data to answer the NEXT 2 QUESTIONS: NO_2 (g) + H_2 O_2 (aq) NO_3 (g) + H_2 O (g); Delta H = -16.15 KJ/mol; K_c = 144 If a scientist observes this reaction at the same temperature starting with 4 M of NO_2 and 4 M of H_2 O_2, what will be the final concentration of NO_3? A. 0.7 M B. 1.8 M C. 3.0 M D. 3.7 M E. 7.4 M Consider the following changes made to the reaction flask in the previous question. Which direction will each of the following changes make the reaction shift? i. increasing temperature ii. increasing the concentration of NO_2 iii. decreasing the volume of the container

Answer 1

Ans. (D)   

NO2(g) + H2O2(l) <===> NO3(g) + H2O(g)

Con: 4 (1-x) 4x 4x

Kc = [NO3][H2O]/[NO2]

144 = 16x² /4(1-x)

4x² +144x -14=0

Using quadratic equation solving

x= 0.9737 M

[NO3] = 4x = 0.9737×4 = 3.8

So answer is D

14)

i) Ans. (A)

Delta H is negative means reaction is exothermic.So increase in temperature shift the reaction to reactant side to nullify the effect of temperature. (Le-Chatlier principle )

ii) B.

increase in [NO2] shift the reaction to product side to nullify the effect of [NO2]. (Le-Chatlier principle )

iii)B (left).

Decrease in volume means increase in no.of moles / unit volume, i.e) pressure increases and reaction proceeds towards backward direction to decrease in no.of moles ,that is towards left.