Which of the following will permanently change the ratio of products to reactant

Question

Which of the following will permanently change the ratio of products to reactants in on equilibrium mixture for a chemical reaction involving gaseous species increasing the temperature adding a catalyst addition of gaseous reactants answers a ond c answers b and c The formation of ammonia from elemental nitrogen and hydrogen is an exothermic process. N_2(g) + 3H_2(g) rightarro 2NH_2(g) delta H = -92.2 KJ/mol Which of the following would drive the reaction to the right addition of ammonia removal of nitrogen an increase in temperature a decrease in pressure none of the above The reaction of nitrogen gas and oxygen gas to form nitrogen monoxide, N_2(g) + O_2(g) rightarrow 2NO(g) has an equilibrium constant of 1.0 x 10^-30 at 298 K. What equilibrium partial pressure of NO(g) will form if 1.0 atm of N_2 and 1.0 atm of O_2 initially are sealed in a flask at 298 K 1.0 x 10^-30 atm 1.0 x 10^-15 atm 1.0 x 10^60 atm 5.0 x 10^-31 atm Nitrosyl bromide decomposes according to the chemical equation below. 2NOBr(g) rightarrow 2NO(g) + Br_2(g) 1.00 atm of NOBr is initially sealed in a flask At equilibrium, the partial pressure of NOBr is 0.82 atm. What is the equilibrium constant for the reaction 3.5 x 10^-2 atm 3.6 x 10^-3 atm 2.8 x 10^-2 atm 8.7 x 10^-3 atm 4.3 times 10^-3 atm

Explanation / Answer

1)

All but adding a catalyst so choose A and B

2)

this is exothermic, menaing it releases hat

none of this will help, all will shift the equilibrium to left

3)

K = NO^2 / (N2*O2)

(10^-30) = (NO^2)/(1*1

NO = sqrt( 10^-30) = 10^-15 answer is B

4)

[NOBR] = 1 - 2x = 0.82

x = (0.82-1)/-2 = 0.09

then

[NO] = 2*0.09= 0.18

[BR2] = 0.09

K = (0.09^2)/(0.09*0.18) = 0.5

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.