How many liters would I need to make a 4M KCl solution using 8 moles of KCI?

A chemist measures the enthalpy change ?H during the following reaction: Fe (s) + 2HCl (g) ? FeCl2 (s) + H2 (g) =?H?157.kJ Use t

erik [133]

Correct Question:

A chemist measures the enthalpy change ΔH during the following reaction: Fe(s) + 2HCl(g)-->FeCl2(s) + H2 ΔH=-157.0 kJ. Use this information to complete the table below. Round each of your answers to the nearest kJ/mol

Answer:

-314 kJ

+628 kJ

+157 kJ

Explanation:

The enthalpy change of a reaction measures the amount of heat that is lost or gained by it. If ΔH >0 the heat is gained, and the reaction is called endothermic, if ΔH<0, the heat is lost, and the reaction is called exothermic.

If the reaction is inverted, the value of ΔH is inverted too (the opposite endothermic reaction is exothermic), and if the reaction is multiplied by a constant, ΔH will be multiplied by it too.

1) 2Fe(s) + 4HCl --> 2FeCl2(s) + 2H2(g)

This reaction is the product of the given reaction by 2, so

ΔH = 2*(-157) = -314 kJ

2) 4FeCl2(s) + 4H2(g) --> 4Fe(s) + 8HCl(g)

This reaction is the inverted reaction given multiplied by 4, so

ΔH = 4*(157) = +628 kJ

3) FeCl2(s) + H2(g) --> Fe(s) + 2HCl

This reaction is the inverted reaction given, so

ΔH = +157 kJ

4 0

3 years ago

Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec

Svetach [21]

The overall reaction is given by:

$Br_{2}(g) + 2 NO(g) \rightarrow 2 NOBr(g)$

The fast step reaction is given as:

$NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})$

The slow step reaction is given as:

$NOBr_{2}(g) + NO(g) \rightarrow 2 NOBr(g)$ (slow step $k_{2}$ )

Now, the expression for the rate of reaction of fast reaction is:

$r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]$

The expression for the rate of reaction of slow reaction is:

$r_{2}=k_{2}[NOBr_{2}] [NO]$

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as $[NOBr_{2}]$ takes place in this reaction.