The decomposition of ammonia is: 2 NH3(g) = N2(g) + 3 H2(g). If Kp is 1.5 × 103 at 400°C, what is the partial pressure of ammonia at equilibrium when N2 is 0.10 atm and H2 is 0.15 atm? 7) A) 4.7 × 10-4 atm B) 2.2 × 10-7 atm C) 2.1 × 103 atm D) 4.4 × 106 atm

The partial pressure of NH3 at the equilibrium is 4.7 * 10^-4 atm (option A )

Explanation:

Step 1: Data given

Kp = 1.5 * 10³ at 400 °C

Partial pressure of N2 = 0.10 atm

Partial pressure of H2 = 0.15 atm

Step 2: The balanced equation

2 NH3(g) → N2(g) + 3 H2(g)

Step 3: Calculate partial pressure of NH3

Kp = (pN2)*(pH2)³ / (pNH3)²

⇒ with Kp = 1.5*10³

⇒with pN2 = the partial pressure of N2 = 0.10 atm

⇒with pH2 = the partial pressure of H2 = 0.15 atm

⇒ with pNH3 = the partial pressure of NH3 = TO BE DETERMINED

1500 = (0.10*0.15³) / (pNH3)²

(pNH3)² = 0.000000225

pNH3 = 0.00047 atm = 4.7 * 10^10^-4 atm

The partial pressure of NH3 at the equilibrium is 4.7 * 10^-4 atm (option A )

more info ? ? i don't understand what the question is

explanation:

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.