What can you determine about the feasibility of a reaction if the enthalpy is positive and

1 answer:

7 0

If the enthalpy is positive and the entropy is positive, the Gibbs energy will always be positive, and the reaction will never be feasible.

<h3>What is the Gibbs Free Energy?</h3>

The Gibb Free Energy is used to obtain the feasibility of a reaction. If the Gibbs free energy is positive the reaction is not spontaneous. If the value is negative, the reaction is spontaneous while a zero values indicates equilibrium.

From the equation;

ΔG = ΔH - TΔS, it follows that if the enthalpy is positive and the entropy is positive, the Gibbs energy will always be positive, and the reaction will never be feasible.

Learn more about Gibbs Free energy:brainly.com/question/20358734

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Formation of water: 2H2 + 1 O2 --> 2H2O

noname [10]

Answer:

2.2 moles H2O

Explanation:

$35g O_2 \mbox{ \cdot }\frac{1mol}{32g/mol} \mbox{ \cdot }\frac{2mol H_2O}{1mol O_2}= 2.1875$ , which rounds to about 2.2

7 0

2 years ago

Natural gas is almost entirely methane. A container with a volume of 2.65L holds 0.120mol of methane. What will the volume be if

mrs_skeptik [129]

The final volume of the methane gas in the container is 6.67 L.

The given parameters;

<em>initial volume of gas in the container, V₁ = 2.65 L</em>
<em>initial number of moles of gas, n₁ = 0.12 mol</em>
<em>additional concentration, n = 0.182 mol</em>

The total number of moles of gas in the container is calculated as follows;

$n_t = 0.12 + 0.182 = 0.302 \ mol$

The final volume of gas in the container is calculated as follows;

$PV = nRT\\\\\frac{V}{n} = \frac{RT}{P} \\\\\frac{V_1}{n_1} = \frac{V_2}{n_2} \\\\V_2 = \frac{V_1 n_2}{n_1} \\\\V_2 = \frac{2.65 \times 0.302}{0.12} \\\\V_2 = 6.67 \ L$