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Kp/Kc for reaction for the equilibrium, A(g) ⇌ C(g)+B(g), is _______.

Sophie [7]

Kp/Kc = RT

<h3>Further explanation</h3>

Given

Reaction

A(g) ⇌ C(g)+B(g)

Required

Kp/Kc

Solution

For reaction :

pA + qB ⇒ mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$

While the equilibrium constant Kp is based on the partial pressure

$\large {\boxed {\bold {Kp ~ = ~ \frac {[pC] ^ m [pD] ^ n} {[pA] ^ p [pB] ^ q}}}}$

The value of Kp and Kc can be linked to the formula '

$\large {\boxed {\bold {Kp ~ = ~ Kc. (R.T) ^ {\Delta n}}}}$

R = gas constant = 0.0821 L.atm / mol.K

Δn=moles products - moles reactants or

number of product coefficients-number of reactant coefficients

For reaction :

A(g) ⇌ C(g)+B(g)

number of product coefficients = 1+1=2

number of reactant coefficients = 1

Δn= 2 - 1 =1

So Kp/Kc = RT

5 0

3 years ago

(You do) If you have 47.2 mol of Na available, along with an excess of Cl₂, how many grams of NaCl can you produce?

IrinaVladis [17]

Answer:

2,760 grams NaCl

Explanation:

To find grams of NaCl, you need to (1) convert moles of Na to moles of NaCl (via mole-to-mole ratio from reaction) and (2) convert moles of NaCl to grams (via molar mass from periodic table). The final answer should have 3 significant figures based on the given measurement.

2 Na + Cl₂ --> 2 NaCl

Molar Mass (NaCl) = 22.99 g/mol + 35.45 g/mol

Molar Mass (NaCl) = 58.44 g/mol

47.2 moles Na 2 moles NaCl 58.44 grams

---------------------- x --------------------------- x ------------------------- =
2 moles Na 1 mole NaCl

= 2,758.368 grams NaCl

= 2,760 grams NaCl

5 0

2 years ago

In the following reaction, what coefficient will be written before the oxygen molecule (O2) to balance the chemical equation? (1

Sergeu [11.5K]

Should be C5H12 + 8O2 yields 5CO2 + 6H2O