Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

a. rate=12Δ[HBr]Δt=−Δ[H2]Δt=−Δ[Br2]Δtrate=12Δ[HBr]Δt=−Δ[H2]Δt=−Δ[Br2]Δt
b. rate=−Δ[HBr]Δt=12Δ[H2]Δt=12Δ[Br2]Δtrate=−Δ[HBr]Δt=12Δ[H2]Δt=12Δ[Br2]Δt
c. rate=−12Δ[HBr]Δt=Δ[H2]Δt=Δ[Br2]Δtrate=−12Δ[HBr]Δt=Δ[H2]Δt=Δ[Br2]Δt
d. rate=Δ[HBr]Δt=−12Δ[H2]Δt=−12Δ[Br2]Δt

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

Thus, the rate is given as:

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

It is conserved.

Explanation: When the box is pushed across the floor, then floor and box becomes warm which resembles the condition of conservation of energy.

The given problem can be resolved by the concept of energy conservation. The statement of energy conservation says that, "Energy is neither be created nor be destroyed, but it is converted from one form to another.

In the given problem, the person pushes the box across the floor, such that the mechanical energy from the moving arms of person is transferred to the box, and the box and floor becomes warm.

During this process, the energy from the person moving arm is transferred to the box in the form of kinetic energy. And thus kinetic energy is then being transferred to thermal energy or heat energy, due to which the box and floor becomes warm.

Thus, we can conclude that the given case reflects the process of transformation of energy from one state to another, which highlights that during the process, energy is conserved.