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Patriots TE Rob Gronkowski named AFC Offensive Player of the Week after dominating the Steelers

Gronkowski had a career-day versus Pittsburgh.

NFL: New England Patriots at Pittsburgh Steelers Charles LeClaire-USA TODAY Sports

The New England Patriots escaped Pittsburgh with a 27-24 victory over the Steelers. The game's best player was a key factor in making the Patriots' come-from-behind win possible: Rob Gronkowski, who the home team was unable to stop and who finished with a career-high 168 receiving yards on nine catches while also adding a critical two-point conversion.

As a result of his dominant performance against the Steelers, the NFL's best tight end has been named the AFC's Offensive Player of the Week. It is the second time that Gronkowski takes home the honor; the first one came in week 14 of the 2011 season, when he registered 160 yards and two touchdowns on six catches against the Washington Redskins.

Gronkowski being named AFC Offensive Player of the Week marks the sixth time this season that a member of the Patriots receives the honor: Quarterback Tom Brady was the conference's offensive player of the week in weeks two, three and 10. Dion Lewis earned the title of the AFC's top special teams player in week 10, Stephen Gostkowski in week 11.