Answer: Adding or removing energy from matter
Explanation:
Answer:
Explanation:
You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.
The formula is
The equation for the reaction is
C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)
ΔS°/J·K⁻¹mol⁻¹ 219.5 205.0 213.6 69.9
The answer is: hydrogen peroxide, H2O2.
H₂O₂(hydrogen peroxide) is pale blue, clear, inorganic liquid.
It is liquid because hydrogen bonds between molecules.
Hydrogen bond is an electrostatic attraction between two polar groups that occurs when a hydrogen atom (H), covalently bound to a highly electronegative atom such as flourine (F), oxygen (O) and nitrogen (N) atoms.
Because of hydrogen bonds, hydrogen peroxide has higher melting and boiling temperatures than other molecules.
Answer:
Mass of C₂H₄N₂ produced = 3.64 g
Explanation:
The balanced chemical equation for the reaction is given below:
3CH₄ (g) + 5CO₂ (g) + 8NH₃ (g) → 4C₂H₄N₂ (g) + 10H₂O (g)
From the equation, 3 moles of CH₄ reacts with 5 moles of CO₂ and 8 moles of NH₃ to produce 4 moles of C₂H₄N₂ and 10 moles of H₂O
Molar masses of the compounds are given below below:
CH₄ = 16 g/mol; CO₂ = 44 g/mol; NH3 = 17 g/mol; C₂H₄N₂ = 56 g/mol; H₂O g/mol
Comparing the mole ratios of the reacting masses;
CH₄ = 1.65/16 = 0.103
CO₂ = 13.5/44 = 0.307
NH₃ = 2.21/17 = 0.130
converting to whole number ratios by dividing with the smallest ratio
CH₄ = 0.103/0.103 = 1
CO₂ = 0.307/0.103 = 3
NH₃ = 0.130/0.103 = 1.3
Multiplying through with 5
CH₄ = 1 × 5 = 5
CO₂ = 3 × 5 = 15
NH₃ = 1.3 × 5 = 6.5
Therefore, the limiting reactant is NH₃
8 × 17 g (136 g) of NH₃ reacts to produce 4 × 56 g (224 g) of C₂H₄N₂
Therefore, 2.21 g of NH₃ will produce (2.21 × 224)/136 g of C₂H₄N₂ = 3.64 g of C₂H₄N₂
Mass of C₂H₄N₂ produced = 3.64 g
Answer:
Thus, the order of the reaction is 2.
The rate constant of the graph which is :- 2.00 M⁻¹s⁻¹
Explanation:
The kinetics of a reaction can be known graphically by plotting the concentration vs time experimental data on a sheet of graph.
The concentration vs time graph of zero order reactions is linear with negative slope.
The concentration vs time graph for a first order reactions is a exponential curve. For first order kinetics the graph between the natural logarithm of the concentration vs time comes out to be a straight graph with negative slope.
The concentration vs time graph for a second order reaction is a hyberbolic curve. Also, for second order kinetics the graph between the reciprocal of the concentration vs time comes out to be a straight graph with positive slope.
Considering the question,
A plot of 1/[NOBr] vs time give a straight line with a slope of 2.00 M⁻¹s⁻¹.
<u>Thus, the order of the reaction is 2.</u>
<u>Also, slope is the rate constant of the graph which is :- 2.00 M⁻¹s⁻¹</u>
