For a chemical reaction with rise in temperature by 10°, the rate constant becomes nearly ____________________.

Correct — C. For most reactions near ordinary temperatures a 10 degree rise nearly doubles the rate constant. That ratio is the temperature coefficient, and option (c) states it.

The reason sits in the Arrhenius equation, k = A e^(-Ea/RT). Temperature appears in a negative exponent, so a small rise in T sharply increases the fraction of molecules carrying at least the activation energy, and k climbs with that fraction.

The growth is exponential, not proportional, which is why ten degrees does so much.

Why the others are wrong

• (a)
  Triple — Triple overstates the rule the stem is quoting. Measured temperature coefficients do run above two for some reactions, but the statement being tested is that the constant is nearly doubled.
• (b)
  half — Half has the direction wrong. Heating raises the share of molecules that clear the activation barrier, so the rate constant rises rather than falls.
• (d)
  one-fourth — One-fourth is the same direction error, only larger. A rate constant does not fall when a reaction is heated, whatever the order of the reaction.

The rate constant k is the temperature-dependent part of a rate law. Concentration terms change the rate but not k.

The Arrhenius equation links k to the activation energy Ea and the absolute temperature T. Two levers raise k: a higher T, which gives more molecules the needed energy, and a lower Ea, which is what a catalyst provides by offering an alternative path.

The doubling for ten degrees is an empirical rule of thumb, not an exact law.

Key facts

• The temperature coefficient is the ratio of rate constants at T+10 and T, and is close to 2 for many reactions.
• The Arrhenius equation is k = A e^(-Ea/RT), where Ea is the activation energy and A the frequency factor.
• A catalyst raises the rate by lowering the activation energy, not by raising the temperature.

Common traps

• Believing the rate constant depends on concentration - it depends on temperature and activation energy.
• Assuming the rise in k with temperature is linear when it is exponential.

No directly related past PYQ was found.