Practice Kinetic Questions

Given:
C(s,graphite) + O₂(g) → CO₂(g) ∆H= -393.5kJ
C(s, diamond) + O₂(g) → CO₂(g) ∆H=-395.4kJ
(Hess' Law)
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Terms in this set (15)
Given:
C(s,graphite) + O₂(g) → CO₂(g) ∆H= -393.5kJ
C(s, diamond) + O₂(g) → CO₂(g) ∆H=-395.4kJ
(Hess' Law)
1. Flip formula #1
CO₂ → C(s, graphite) + O₂ ∆H=+393.5kJ
C(s, diamond) + O₂(g) → CO₂(g) ∆H=-395.4kJ
2. Cancel out like terms
3. Add ∆H
-395.4+393.5= -1.9kJ
Given:
H₂SO₄(l) → SO₃(g) + H₂O(g)
H₂S(g) + 2O₂(g) → H₂SO₄(l) ∆H=-235.5 kJ
H₂S(g) + 2O₂(g)→ SO₃(g) + H₂O(l) ∆H=-207 kJ
H₂O(l) → H₂O(g) ∆H=44 kJ
(Hess' Law)
1. Flip formula #2
H₂SO₄(l) → SO₃(g) + H₂O(g)
H₂SO₄(l) → H₂S(g) + 2O₂(g) ∆H=235.5 kJ
H₂S(g) + 2O₂(g)→ SO₃(g) + H₂O(l) ∆H=-207 kJ
H₂O(l) → H₂O(g) ∆H=44 kJ
2. Cancel out like terms
3. Add ∆H
235.5-207+44= 72.5kJ
Given:
HCl(g) + NaNO₂(s) → HNO₂(l) + NaCl(s)
2NaCl + H₂O(l) → 2HCl(g) + Na₂O(s) ∆H=507 kJ
NO(g) + NO₂(g) + Na₂O(s) → 2NaNO₂(s) ∆H=-427 kJ
NO(g) + NO₂(g) → N₂O(g) + O₂(g) ∆H=-43 kJ
2HNO₂(l) → N₂O(g) + O₂(g) + H₂O(l) ∆H= 34 kJ
(Hess' Law)
1. Flip formulas #2, #3, and #5
HCl(g) + NaNO₂(s) → HNO₂(l) + NaCl(s)
2HCl(g) + Na₂O(s) → 2NaCl + H₂O(l) ∆H=-507 kJ
2NaNO₂(s) → NO(g) + NO₂(g) + Na₂O(s) ∆H=427 kJ
NO(g) + NO₂(g) → N₂O(g) + O₂(g) ∆H=43 kJ
N₂O(g) + O₂(g) + H₂O(l) → 2HNO₂(l) ∆H=-34 kJ
2. Divide formulas #2, #3, #4, and #5 by 2
HCl(g) + NaNO₂(s) → HNO₂(l) + NaCl(s)
HCl(g) + 1/2 Na₂O(s)→ NaCl + 1/2 H₂O(l) ∆H=-507/2 kJ
NaNO₂(s) → 1/2 NO(g) + 1/2 NO₂(g) + 1/2 Na₂O(s)
∆H=427/2 kJ
1/2 NO(g) + 1/2 NO₂(g) → 1/2 N₂O(g) + 1/2 O₂(g)
∆H=43/2 kJ
1/2 N₂O(g) + 1/2 O₂(g) + 1/2 H₂O(l) → HNO₂(l)
∆H=-34/2 kJ
3. Cancel out like terms
4. Add ∆H
-507/2 + 427/2 + 43/2 -34/2 =78.5kJ
Calculate the equilibrium constant.
N₂(g) + 3H₂(g) ↔ 2NH₃(g)
K= [NH₃]² / [N₂] [H₂]²
Calculate the equilibrium constant.
Li₂CO₃(s) → 2Li⁺(aq) + CO₃⁻²(aq)
K= [Li]² [CO₃⁻²]
Calculate the equilibrium constant of
K=[NH₃]⁻² / [N₂] [H₂]²
when [NH₃]= .01 M , [N₂]= .2 M , [H₂]= .02 M
K= .01² / .02(.02)³
K=625
Calculate the equilibrium constant of
K=[H] [OH]
when [H]= 1 x 10⁻⁸ M , [OH]= 1 x 10⁻⁶ M
K= [1 x 10⁻⁸] [1 x 10⁻⁶]
K= .01 x 10⁻¹⁴
In the equation
N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal
if more N₂ is added, what happens to
a. equilibrium shift?
b. N₂ ?
c. H₂ ?
d. NH₃ ?
a. shift right
b. decrease
c. decrease
d. increase
In the equation
N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal
if more NH₃ is added, what happens to
a. equilibrium shift?
b. N₂ ?
c. H₂ ?
d. NH₃ ?
a. shift left
b. increase
c. increase
d. decrease
In the equation
N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal
if N₂ is removed, what happens to
a. equilibrium shift?
b. N₂ ?
c. H₂ ?
d. NH₃ ?
a. shift left
b. increase
c. increase
d. decrease
In the equation N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal if more NH₃ is removed, what happens to a. equilibrium shift? b. N₂ ? c. H₂ ? d. NH₃ ?a. shift right b. decrease c. decrease d. increaseIn the equation N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal if temperature is increased, what happens to a. equilibrium shift? b. N₂ ? c. H₂ ? d. NH₃ ?a. shift left b. increase c. increase d. decreaseIn the equation N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal if temperature is decreased, what happens to a. equilibrium shift? b. N₂ ? c. H₂ ? d. NH₃ ?a. shift right b. decrease c. decrease d. increaseIn the equation N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal if pressure is increased, what happens to a. equilibrium shift? b. N₂ ? c. H₂ ? d. NH₃ ?a. shift right b. decrease c. decrease d. increaseIn the equation N₂(g) + 3H₂(g) ↔ 2NH₃(g) +22.0 kcal if pressure is decreased, what happens to a. equilibrium shift? b. N₂ ? c. H₂ ? d. NH₃ ?a. shift left b. increase c. increase d. decrease