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3 years ago

Which is the correct expression for Keq when the chemical reaction is as follows:

2 answers:

4 0

The expression of the Keq is [NH3]2/([N2][H2]3). The common formula of reaction: ax+by=cz is Keq - [z]c/([x]a*[y]b). Remember to use a balanced chemical reaction.

skelet666 [1.2K]3 years ago

3 0

Answer : The expression for equilibrium constant will be,

$K_{eq}=\fac{[NH_3]^2}{[N_2][H_2]^3}$

Explanation :

Equilibrium constant : It is defined as the number of ratio of the concentration between the products and the reactants at equilibrium.

The equilibrium reaction is,

$aX+bY\rightleftharpoons cW+dZ$

The general expression for equilibrium constant will be,

$K_{eq}=\frac{[X]^a[Y]^b}{[W]^c[Z]^d}$

The given balanced chemical reaction will be,

$N_2+3H_2\rightleftharpoons 2NH_3$

The expression for equilibrium constant will be,

$K_{eq}=\frac{[NH_3]^2}{[N_2][H_2]^3}$

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If 5.0 liters H2 (g) at STP is heated to a temperature of 985, pressure remaining constant, the new volume of the gas will be?

vlada-n [284]

Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

Chemistry - Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

Step 1: Define

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

Step 2: Solve for new volume

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
Cross-multiply: $(5 \ L \ H_2)(985 \ K) = (x \ L \ H_2)(273 \ K)$
Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate x: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

Step 3: Check

We are given 2 sig figs as the smallest. Follow sig fig rules and round.

$18.0403 \ L \ H_2 \approx 18 \ L \ H_2$

5 0

3 years ago

How many grams of water will be produced from 50 g hydrogen reacting with 50 g oxygen?

DanielleElmas [232]

We need to first come up with a balanced equation:

$4H+O_{2}$ → $2H_{2}O$

We know that the molar ratio of hydrogen to oxygen to water now is 4:1:2.

Converting the amount of grams given to moles is as follows:

Hydrogen: $\frac{1mole}{1.008g}*50g=49.6mol$

Oxygen: $\frac{1mole}{15.999g}*50g=3.125mol$

We know now that the limiting reactant is oxygen. We can then know that the number of moles of water are produced are double the number of moles of oxygen used due to the ratio that we established at the beginning - 4:1:2.

So we now can use 6.25 moles of water as the amount produced.

Then we convert moles of water to grams:

$\frac{18.015g}{1mole} *6.25mol=112.59g$

Now we know that there are 112.59g of water produced when we start with 50g of hydrogen and 50g of water.

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3 years ago

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Masja [62]

Answer:

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polet [3.4K]

Answer:

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3 years ago

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Stolb23 [73]

Answer:

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Explanation:

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3 years ago